Chapter 1: Geometric Twins

To recreate the symbol on another board, we need a way to produce an exact copy of the figure. A very direct method is to trace the outline of the symbol using tracing paper and then transfer this tracing to the new board.

Place a sheet of tracing paper on top of the given symbol, carefully copy the outline with a pencil, and then paste or press this tracing onto the new board so that the outline can be inked over.

This tracing method reproduces the figure perfectly, but it works only when the symbol is small enough to fit on a sheet of tracing paper. For large symbols we cannot use tracing paper directly, so we look for another approach — measure a few key parts of the symbol and use those measurements to redraw it accurately.

Yes. The symbol is made up of two straight arms that meet at a corner. If we label the corner points as $$A$$, $$B$$, $$C$$ (with $$B$$ at the meeting point of the two arms), then the shape is completely described by:

• the length of the first arm $$AB$$,
• the length of the second arm $$BC$$, and
• the angle $$\angle ABC$$ between the two arms.

Knowing these three quantities, we can start from any point, mark $$B$$, draw a segment of length $$AB$$ in one direction to fix $$A$$, turn through the angle $$\angle ABC$$, and draw a segment of length $$BC$$ to fix $$C$$. This gives an exact copy of the original tick-mark symbol.

No. The two arm lengths alone are not sufficient.

Take $$AB = 4\,\mathrm{cm}$$ and $$BC = 8\,\mathrm{cm}$$. Starting from the point $$B$$, we can draw a segment of length $$4\,\mathrm{cm}$$ to fix $$A$$, but the second segment $$BC$$ of length $$8\,\mathrm{cm}$$ can be drawn in any direction from $$B$$. Each different direction gives a different opening between the two arms, and hence a different-looking symbol.

The four sample figures in the textbook show exactly this: the arm lengths are the same in each, yet the opening angle $$\angle ABC$$ changes, and so the shapes are different. Hence the two arm lengths by themselves do not fix the figure.

Yes. The one extra measurement that we still need is the angle between the two arms, namely $$\angle ABC$$.

As soon as we know $$\angle ABC$$ in addition to the arm lengths $$AB$$ and $$BC$$, the shape and size of the symbol are completely fixed. There is no more freedom left to draw a different figure.

So the three measurements — the arm length $$AB$$, the arm length $$BC$$, and the included angle $$\angle ABC$$ — together give an exact replica.

Yes, these three measurements are enough. Here is a step-by-step construction.

Step 1. Mark a point $$B$$ on the paper.

Step 2. Using a ruler, draw a segment from $$B$$ of length $$4\,\mathrm{cm}$$ and mark its other end as $$A$$. So $$AB = 4\,\mathrm{cm}$$.

Step 3. Place the centre of a protractor at $$B$$ and its base along $$BA$$. From the $$0^{\circ}$$ mark on the ray $$BA$$, count round to $$80^{\circ}$$ and mark a small dot. Draw a ray from $$B$$ through this dot; this ray makes $$\angle ABC = 80^{\circ}$$ with $$BA$$.

Step 4. On this new ray, use a ruler to mark the point $$C$$ so that $$BC = 8\,\mathrm{cm}$$.

The two segments $$BA$$ and $$BC$$ together form the required tick-mark symbol with $$AB = 4\,\mathrm{cm}$$, $$BC = 8\,\mathrm{cm}$$ and $$\angle ABC = 80^{\circ}$$.

No. Having equal arm lengths $$AB$$ and $$BC$$ is not by itself enough to conclude that two such symbols are congruent.

We have already seen that, with the arms fixed at $$AB = 4\,\mathrm{cm}$$ and $$BC = 8\,\mathrm{cm}$$, one can draw many different-looking symbols simply by changing the angle at $$B$$. Two such symbols with different angles at $$B$$ cannot be made to fit exactly on top of each other, so they are not congruent.

The two symbols will be congruent only when, in addition to $$AB$$ and $$BC$$ being equal, the included angle $$\angle ABC$$ is also equal in the two figures.

Each of the two figures is a tick-mark type symbol formed by two straight arms meeting at a corner. Label the two arm-lengths of the first figure as $$AB$$ and $$BC$$ (with the corner at $$B$$) and the corresponding lengths of the second figure as $$A'B'$$ and $$B'C'$$.

Two such figures are congruent only when

• the first arms are equal: $$AB = A'B'$$,
• the second arms are equal: $$BC = B'C'$$, and
• the included angles at the corners are equal: $$\angle ABC = \angle A'B'C'$$.

Measure the two arms and the corner-angle of each figure with a ruler and protractor. On measuring, the two figures in the book have the same arm lengths and the same angle between the arms. So one can be placed exactly over the other (possibly after flipping) — they fit exactly.

Hence the two figures are congruent.

Two figures are congruent when one can be placed exactly on the other, if necessary after rotating or flipping it. So we compare the two figures of each pair by eye (and, more carefully, by tracing paper).

• Pair 1 — the two water-drops. They have the same shape and the same size; one drop is essentially a mirror image of the other. On tracing one and flipping it, it fits exactly on the second. Congruent.
• Pair 2 — the two clouds. One cloud is clearly larger than the other; a tracing of the small cloud would not cover the big one. Not congruent.
• Pair 3 — the two star-bursts. They have the same general spiky shape, but one is bigger than the other and the number/size of the points is not identical. Not congruent.
• Pair 4 — the two leaves. The two leaves have the same outline and the same size (one is just tilted differently). Rotating the tracing of one makes it fit exactly over the other. Congruent.

(a) Circle. A circle is fixed as soon as we know how big it is. The size of a circle is completely described by a single measurement — its radius $$r$$ (or equivalently its diameter $$d = 2r$$). Given the radius, we can put the compass point anywhere on the paper, open the compass to length $$r$$, and draw a circle congruent to the given one.

So to check whether two circles are congruent, we only need to measure the radius (or diameter) of each. If the two radii are equal, the two circles are congruent. If the radii are unequal, they are not.

(b) Rectangle. A rectangle already has all four angles equal to $$90^{\circ}$$. So its shape and size are decided by just two measurements — its length $$l$$ and its breadth $$b$$. Given $$l$$ and $$b$$, we can construct a congruent rectangle: draw the base of length $$l$$, at each end draw a perpendicular of length $$b$$, and join the tops.

So to check whether two rectangles are congruent, measure the length and the breadth of each. If both dimensions match — the length of one equals the length of the other and the breadth of one equals the breadth of the other — then the two rectangles are congruent. Otherwise they are not.

The figure shown is made of three straight arms meeting at one common point (a "Y"-shaped figure). Label the common meeting point as $$O$$, and the three end-points of the arms as $$P$$, $$Q$$, $$R$$.

Such a figure is completely fixed once we know

• the three arm lengths $$OP$$, $$OQ$$, $$OR$$, and
• the angles between consecutive arms at the vertex $$O$$: $$\angle POQ$$, $$\angle QOR$$, $$\angle ROP$$.

(Two of these three angles are enough, because the three angles around $$O$$ add up to $$360^{\circ}$$.)

To check whether two Y-figures are congruent, measure the three arm lengths and the angles between the arms in both. If the corresponding arm lengths match and the corresponding angles match (allowing for a possible rotation or flip), the two figures are congruent — a tracing of one will fit exactly on the other.

Pair 1. Compare the two given Y-figures by tracing one on tracing paper and trying to place it on the other. The corresponding arm lengths and the corresponding angles between the arms are equal, so one fits exactly on the other. Hence the two figures in Pair 1 are congruent.

Pair 2. Comparing the two Y-figures in the second pair, the arms of one figure are longer than the arms of the other (or the angles between the arms differ). A tracing of one does not fit on the other. Hence the two figures in Pair 2 are not congruent.

Meera and Rabia want to make a cardboard cutout that is congruent to the triangular frame in school. The frame is too big to trace, so tracing paper is ruled out.

What they can do is measure a few important parts of the triangular frame and then use those measurements to construct an identical triangle on cardboard. In particular, they can use a measuring tape to record the lengths of the three sides of the triangular frame (and, if needed, a protractor for the angles). Once these measurements are known, a triangle with exactly the same shape and size can be constructed anywhere, including on the cardboard.

So the sensible plan is: measure the sides (and if required the angles) of the frame, and construct a triangle with the same measurements on the cardboard.

Meera claims that once the three side-lengths of the triangle are known, we do not need to measure the angles — the sides alone are enough to make a triangle congruent to the given one.

Yes, Meera is right. When the three side-lengths of a triangle are fixed, the shape of the triangle is also fixed — the angles are automatically determined. This is the SSS (Side–Side–Side) condition for congruence: if the three sides of one triangle equal the three sides of another triangle, the two triangles are congruent.

We can convince ourselves of this by construction: with the three sides given, only one triangle (up to a flip) can be built by the standard ruler-and-compass method — draw one side as the base, and from its two ends draw circular arcs whose radii are the other two sides; the arcs meet in exactly one point on each side of the base, giving essentially one triangle.

With side-lengths $$4\,\mathrm{cm}$$, $$6\,\mathrm{cm}$$ and $$8\,\mathrm{cm}$$, the triangle is small enough to construct on a page. Here is how we do it.

Step 1. Draw a segment $$AB = 6\,\mathrm{cm}$$ as the base.

Step 2. With centre $$A$$ and radius $$4\,\mathrm{cm}$$, draw an arc above $$AB$$.

Step 3. With centre $$B$$ and radius $$8\,\mathrm{cm}$$, draw another arc that cuts the first arc. Call the point of intersection $$E$$.

Step 4. Join $$AE$$ and $$BE$$. Then $$\triangle ABE$$ has $$AB = 6\,\mathrm{cm}$$, $$AE = 4\,\mathrm{cm}$$, $$BE = 8\,\mathrm{cm}$$ — the required triangle.

If, instead, we let the two arcs cut each other below the base $$AB$$ at a point $$F$$, then $$\triangle ABF$$ also has $$AB = 6\,\mathrm{cm}$$, $$AF = 4\,\mathrm{cm}$$, $$BF = 8\,\mathrm{cm}$$. But $$\triangle ABF$$ is just the mirror image of $$\triangle ABE$$ about the line $$AB$$, so the two triangles are congruent — the construction gives essentially one triangle. Hence the three side-lengths determine a unique triangle.

Yes. Just the three side-lengths $$4\,\mathrm{cm}$$, $$6\,\mathrm{cm}$$, $$8\,\mathrm{cm}$$ are enough to produce a triangle of exactly the same shape and size — this is the SSS criterion.

Construction.

Step 1. Draw the base $$AB$$ of length $$6\,\mathrm{cm}$$.

Step 2. With centre $$A$$ and radius $$4\,\mathrm{cm}$$, draw an arc.

Step 3. With centre $$B$$ and radius $$8\,\mathrm{cm}$$, draw a second arc, cutting the first arc at a point $$C$$.

Step 4. Join $$AC$$ and $$BC$$.

The resulting $$\triangle ABC$$ has $$AB = 6\,\mathrm{cm}$$, $$CA = 4\,\mathrm{cm}$$, $$BC = 8\,\mathrm{cm}$$ — a triangle with exactly the given side-lengths, and hence congruent to the given triangle.

In the construction, both $$E$$ and $$F$$ are points at which the arc of radius $$4\,\mathrm{cm}$$ (centred at $$A$$) meets the arc of radius $$8\,\mathrm{cm}$$ (centred at $$B$$). The point $$E$$ lies on one side of the base $$AB$$ and the point $$F$$ lies on the other side.

Compare the two triangles $$\triangle ABE$$ and $$\triangle ABF$$:

$$AB = AB \quad (\text{common side})$$

$$AE = AF = 4\,\mathrm{cm} \quad (\text{radii of the arc drawn from } A)$$

$$BE = BF = 8\,\mathrm{cm} \quad (\text{radii of the arc drawn from } B)$$

All three pairs of corresponding sides are equal. By the SSS congruence criterion, we have

$$\triangle ABE \cong \triangle ABF.$$

Geometrically, $$\triangle ABF$$ is the mirror image (reflection) of $$\triangle ABE$$ about the line $$AB$$. Flipping the triangle across $$AB$$ makes it fit exactly on the other triangle, confirming congruence.

From the tick-marks in the figure, the equal sides in $$\triangle ABC$$ and $$\triangle XYZ$$ are

$$AB = YZ \;(\text{single tick})$$,
$$BC = XY \;(\text{double tick})$$,
$$CA = XZ \;(\text{triple tick})$$.

For the two triangles to fit exactly one over the other, each side of $$\triangle ABC$$ must lie on the equal side of $$\triangle XYZ$$. So we must overlap the endpoints of the equal sides.

Look at the side $$AB$$ (single tick). It is equal to $$YZ$$ (single tick). So the pair $$\{A, B\}$$ must be placed on the pair $$\{Y, Z\}$$. In particular, $$B$$ is the endpoint that $$AB$$ shares with $$BC$$ (double tick), and the side equal to $$BC$$ (i.e. $$XY$$) meets $$YZ$$ at $$Y$$. Hence $$B$$ must go on $$Y$$, and therefore $$A$$ must go on $$Z$$.

Similarly, $$C$$ is the vertex shared by $$BC$$ (double tick) and $$CA$$ (triple tick). The corresponding vertex in $$\triangle XYZ$$ is $$X$$, since $$X$$ is shared by $$XY$$ (double) and $$XZ$$ (triple). So $$C$$ goes on $$X$$.

The correct correspondence is therefore

$$A \leftrightarrow Z,\quad B \leftrightarrow Y,\quad C \leftrightarrow X,$$

and we may write

$$\triangle ABC \cong \triangle ZYX.$$

With this matching, every pair of equal sides overlaps exactly and the two triangles superimpose.

To make the triangles fit exactly, the sides that overlap must be equal, i.e., each pair of sides that lie on top of each other must have the same number of ticks. In the given triangles the three sides carry different markings (one single, one double, one triple), so a side of $$\triangle ABC$$ can be laid only on the one side of $$\triangle XYZ$$ that has the same marking. This forces the vertices to correspond in exactly one way.

Working this out (as in the previous question) we get

$$A \leftrightarrow Z,\; B \leftrightarrow Y,\; C \leftrightarrow X.$$

Any other pairing of the vertices would lay a side with (say) a single tick over a side with a double tick — and since those sides are of different lengths, the two triangles would not fit.

So there is no other way of overlapping the vertices: the correspondence $$\triangle ABC \cong \triangle ZYX$$ is the only one that makes the triangles fit exactly.

The figure shows a rectangle $$ABCD$$ with the diagonal $$BD$$ drawn, dividing the rectangle into two triangles $$\triangle ABD$$ and $$\triangle CDB$$.

Since $$ABCD$$ is a rectangle, the opposite sides are equal:

$$AB = CD, \qquad AD = CB.$$

Also, the two triangles share the diagonal $$BD$$:

$$BD = BD \quad (\text{common side}).$$

So in $$\triangle ABD$$ and $$\triangle CDB$$ the three pairs of corresponding sides are equal:

$$AB = CD,\quad AD = CB,\quad BD = DB.$$

By the SSS congruence criterion,

$$\triangle ABD \cong \triangle CDB.$$

Note that the correspondence of vertices is $$A \leftrightarrow C$$, $$B \leftrightarrow D$$, $$D \leftrightarrow B$$; the side $$AB$$ overlaps $$CD$$, the side $$AD$$ overlaps $$CB$$, and the diagonal $$BD$$ overlaps $$DB$$ (itself, read in the opposite direction).

Take a paper rectangle $$ABCD$$ of the same shape as the one in Fig. 1.1 and cut it along its diagonal $$BD$$. This gives two triangular cutouts: $$\triangle ABD$$ and $$\triangle CDB$$.

Now try to place $$\triangle ABD$$ on $$\triangle CDB$$ so that the vertex $$A$$ lies on $$C$$, and $$B$$ and $$D$$ interchange. To do this we need to flip $$\triangle ABD$$ over (turn it upside down) and then place it on $$\triangle CDB$$. After this flip:

• Side $$AB$$ (of $$\triangle ABD$$) lies exactly on side $$CD$$ (of $$\triangle CDB$$), since $$AB = CD$$.
• Side $$AD$$ lies exactly on side $$CB$$, since $$AD = CB$$.
• Side $$BD$$ lies exactly on side $$DB$$ — the common diagonal.

The two cutouts fit exactly one over the other, verifying that $$\triangle ABD \cong \triangle CDB$$.

(Note that if we try to place $$A$$ on $$C$$, $$B$$ on $$B$$ and $$D$$ on $$D$$ — i.e. use the correspondence $$\triangle ABD \leftrightarrow \triangle CDB$$ with $$A \to C, B \to B, D \to D$$ — the side $$AB$$ would try to lie on $$CB$$, but these need not be equal, so this superposition does not work. Only the correspondence $$A \to C, B \to D, D \to B$$ gives a perfect fit.)

The two triangles are $$\triangle ABD$$ (formed by vertices $$A$$, $$B$$, $$D$$ of the rectangle) and $$\triangle CDB$$ (formed by $$C$$, $$D$$, $$B$$), obtained from the rectangle $$ABCD$$ by drawing the diagonal $$BD$$.

The corresponding sides must be the equal pairs. In the rectangle:

$$AB = CD,\quad AD = CB,\quad BD = DB\;(\text{common}).$$

Matching sides of equal length pairs up the vertices as follows:

• $$AB$$ (in $$\triangle ABD$$) matches $$CD$$ (in $$\triangle CDB$$), giving $$A \leftrightarrow C$$ and $$B \leftrightarrow D$$.
• $$AD$$ matches $$CB$$, giving $$A \leftrightarrow C$$ and $$D \leftrightarrow B$$.
• $$BD$$ matches $$DB$$, consistent with $$B \leftrightarrow D$$ and $$D \leftrightarrow B$$.

So the correct correspondence is

$$A \leftrightarrow C,\quad B \leftrightarrow D,\quad D \leftrightarrow B.$$

Writing the vertices of the two triangles in matching order gives the congruence

\[ \triangle ABD \cong \triangle CDB. \]

The statement $$\triangle HEN \cong \triangle BIG$$ tells us the correspondence of vertices:

$$H \leftrightarrow B,\quad E \leftrightarrow I,\quad N \leftrightarrow G.$$

Any other correct way of writing this congruence must list the vertices of the two triangles in a matching order — if we shuffle the vertices of the first triangle, we must apply the same shuffle to the vertices of the second. A triangle has $$3! = 6$$ orderings of its vertices, so there are $$6$$ correct ways of writing the same congruence (the given one plus $$5$$ others).

Applying each of the $$6$$ orderings to $$H, E, N$$ (and the same ordering to $$B, I, G$$) we get:

1. $$\triangle HEN \cong \triangle BIG$$ (the given one)
2. $$\triangle HNE \cong \triangle BGI$$
3. $$\triangle EHN \cong \triangle IBG$$
4. $$\triangle ENH \cong \triangle IGB$$
5. $$\triangle NHE \cong \triangle GBI$$
6. $$\triangle NEH \cong \triangle GIB$$

So, apart from the given one, the other five correct ways are the ones listed in items $$2$$–$$6$$.

From the figure, the sides of $$\triangle RED$$ are

$$RE = 3.5\,\mathrm{cm},\quad ED = 5\,\mathrm{cm},\quad DR = 6\,\mathrm{cm},$$

and the sides of $$\triangle JAM$$ (labelled as $$J$$, $$A$$, $$M$$) are

$$JA = 3.5\,\mathrm{cm},\quad AM = 5\,\mathrm{cm},\quad MJ = 6\,\mathrm{cm}.$$

The three side-lengths of the two triangles are the same set $$\{3.5\,\mathrm{cm}, 5\,\mathrm{cm}, 6\,\mathrm{cm}\}$$, so the triangles satisfy the SSS congruence criterion.

To express the congruence, match equal sides:

• $$RE = JA = 3.5\,\mathrm{cm}\;\;\Rightarrow\;\; R \leftrightarrow J,\; E \leftrightarrow A.$$
• $$ED = AM = 5\,\mathrm{cm}\;\;\Rightarrow\;\; E \leftrightarrow A,\; D \leftrightarrow M.$$
• $$DR = MJ = 6\,\mathrm{cm}\;\;\Rightarrow\;\; D \leftrightarrow M,\; R \leftrightarrow J.$$

All three match consistently. Hence

\[ \triangle RED \cong \triangle JAM. \]

The dashed segment $$AC$$ divides the given kite-shaped figure into two triangles $$\triangle ABC$$ and $$\triangle ADC$$.

Congruence. Compare $$\triangle ABC$$ and $$\triangle ADC$$:

$$AB = AD \quad (\text{given}),$$

$$CB = CD \quad (\text{given}),$$

$$AC = AC \quad (\text{common side}).$$

All three pairs of corresponding sides are equal, so by the SSS criterion,

\[ \triangle ABC \cong \triangle ADC. \]

Does $$AC$$ bisect $$\angle BAD$$ and $$\angle BCD$$?

Since $$\triangle ABC \cong \triangle ADC$$, the corresponding angles of these two congruent triangles are equal. In particular:

$$\angle BAC = \angle DAC \quad (\text{corresponding angles at vertex } A),$$

$$\angle BCA = \angle DCA \quad (\text{corresponding angles at vertex } C).$$

But $$\angle BAC$$ and $$\angle DAC$$ are the two parts into which $$AC$$ divides $$\angle BAD$$, and $$\angle BCA$$, $$\angle DCA$$ are the two parts into which $$AC$$ divides $$\angle BCD$$. Since the two parts are equal in each case, $$AC$$ bisects both $$\angle BAD$$ and $$\angle BCD$$.

Compare $$\triangle DFE$$ and $$\triangle GED$$ side by side, in the order in which the vertices are written.

• First pair: $$DF$$ (in $$\triangle DFE$$) and $$GE$$ (in $$\triangle GED$$). Are these equal?
  We are given $$DF = DG$$ and $$FE = GE$$. So $$GE = FE$$, but there is no reason to say $$GE = DF$$. In general, $$DF \ne GE$$.
• Second pair: $$FE$$ and $$ED$$. We are given $$FE = GE$$, but not $$FE = ED$$.
• Third pair: $$ED$$ and $$DG$$. Not given equal.

So the correspondence $$D \leftrightarrow G,\; F \leftrightarrow E,\; E \leftrightarrow D$$ (which the statement "$$\triangle DFE \cong \triangle GED$$" would demand) does not match sides of equal length.

The correct congruence is a different one. In fact, comparing $$\triangle DFE$$ with $$\triangle DGE$$ (note the change in order):

$$DF = DG \quad (\text{given}),$$

$$FE = GE \quad (\text{given}),$$

$$DE = DE \quad (\text{common side}).$$

By the SSS criterion, $$\triangle DFE \cong \triangle DGE$$.

So the two triangles are congruent, but the correct way of writing this congruence is $$\triangle DFE \cong \triangle DGE$$, not $$\triangle DFE \cong \triangle GED$$ (the vertex order given in the question is incorrect).

Check: $$30^{\circ} + 70^{\circ} + 80^{\circ} = 180^{\circ}$$, so these three angles can indeed be the angles of a triangle.

But knowing the three angles alone does not fix the size of the triangle. For any triangle with angles $$30^{\circ}$$, $$70^{\circ}$$ and $$80^{\circ}$$, we can shrink it or enlarge it and still keep the angles the same. The book actually shows three such triangles: they all have the given three angles, but they are of different sizes and are clearly not congruent.

Since we need the copy to be exactly the same size as the original frame, three angles alone are not sufficient. To fix the size we must also know at least one side-length.

Hence, using only the three angles $$30^{\circ},\,70^{\circ},\,80^{\circ}$$, we cannot create an exact copy of the frame.

In $$\triangle ABC$$, the angle $$\angle A$$ is the angle between the two sides $$AB$$ and $$AC$$ (the sides meeting at vertex $$A$$). Similarly, in $$\triangle XYZ$$, the angle $$\angle X$$ lies between $$XY$$ and $$XZ$$.

So we are given the following equal pairs:

• Two sides: $$AB = XY = 6\,\mathrm{cm}$$ and $$AC = XZ = 5\,\mathrm{cm}$$.
• The angle included between these two sides: $$\angle A = \angle X = 30^{\circ}$$.

This is exactly the SAS (Side–Angle–Side) condition: two sides and the included angle of one triangle are equal to two sides and the included angle of the other.

The SAS condition guarantees congruence. Hence

\[ \triangle ABC \cong \triangle XYZ. \]

We are asked to construct a triangle in which two sides and the included angle are $$6\,\mathrm{cm},\; 5\,\mathrm{cm},\; 30^{\circ}$$.

Step 1. Draw a segment $$AB$$ of length $$6\,\mathrm{cm}$$ as the base.

Step 2. Place the centre of a protractor at $$A$$ with its base line along $$AB$$. Mark a point above $$AB$$ at $$30^{\circ}$$ from $$AB$$, and draw a ray from $$A$$ through this point. This ray makes $$\angle BAC = 30^{\circ}$$.

Step 3. On this ray, mark the point $$C$$ such that $$AC = 5\,\mathrm{cm}$$.

Step 4. Join $$BC$$.

The resulting $$\triangle ABC$$ has $$AB = 6\,\mathrm{cm}$$, $$AC = 5\,\mathrm{cm}$$ and $$\angle A = 30^{\circ}$$. On comparing constructions made by different classmates (each starting with a fresh sheet), we notice that every triangle we construct in this way is congruent to every other — they all fit exactly. This shows that the SAS condition determines a unique triangle.

Look at where the given angle sits with respect to the given sides.

• In $$\triangle ABC$$: $$\angle B$$ is at vertex $$B$$. The side $$AB$$ has $$B$$ as an endpoint, so $$\angle B$$ touches $$AB$$. But the second given side $$AC$$ does not have $$B$$ as an endpoint — so $$\angle B$$ is not between $$AB$$ and $$AC$$; it is a non-included angle.
• In $$\triangle XYZ$$: similarly, $$\angle Y$$ touches $$XY$$ but not $$XZ$$.

So the data is of the form Side–Side–Angle (SSA), with the angle not between the two given sides.

The SSA condition does not guarantee congruence. In general, two triangles satisfying only SSA may or may not be congruent — we cannot conclude congruence from this data alone.

(The next problem asks us to verify this by actually constructing triangles from these measurements.)

Yes — the construction shows that two non-congruent triangles can have the same $$AB, AC, \angle B$$ measurements. Let us verify.

Rough diagram. Rename the triangle $$\triangle PQR$$ with $$PQ = AB = 6\,\mathrm{cm}$$, $$\angle Q = \angle B = 30^{\circ}$$ (at vertex $$Q$$), and $$PR = AC = 4\,\mathrm{cm}$$ (the side opposite $$Q$$ is not fixed — only its length from $$P$$ is).

Step 1. Draw the base $$PQ = 6\,\mathrm{cm}$$.

Step 2. Draw a ray from $$Q$$ making an angle $$30^{\circ}$$ with $$QP$$; call this ray $$l$$. The third vertex $$R$$ lies on $$l$$.

Step 3. With centre $$P$$ and radius $$4\,\mathrm{cm}$$, draw an arc. This arc cuts the ray $$l$$ at two different points — call them $$R$$ and $$S$$.

Step 4. Both $$\triangle PQR$$ and $$\triangle PQS$$ have

$$PQ = 6\,\mathrm{cm},\quad \angle Q = 30^{\circ},\quad PR = PS = 4\,\mathrm{cm}.$$

Yet $$\triangle PQR$$ and $$\triangle PQS$$ are clearly of different shapes and sizes — one is much smaller than the other.

So two non-congruent triangles can indeed satisfy the given SSA measurements, confirming that SSA is not a valid congruence condition.

In the construction (Step 3), the arc from $$Q$$ of radius $$4\,\mathrm{cm}$$ cuts the line $$l$$ at two points, marked $$R$$ and $$S$$. Joining these to $$P$$ (in the book's notation) or to $$P$$ and $$Q$$ appropriately gives two triangles $$\triangle PQR$$ and $$\triangle PQS$$.

Both triangles satisfy the given measurements $$PQ = 6\,\mathrm{cm}$$, $$\angle Q = 30^{\circ}$$ and (side from $$Q$$ to the vertex on $$l$$) $$= 4\,\mathrm{cm}$$. So the figure yields two possible triangles, not one.

Hence there is no single "required triangle": the SSA data can be completed to a triangle in two different ways, and both are valid. The figure shows that these two triangles differ in size and shape, which means the given measurements do not fix a unique triangle.

So the answer is: both intersection points of the arc with the line give valid triangles, so we get two different triangles from the figure, not one.

Look at how the given angle-pairs relate to the given side $$BC$$.

• $$\angle B$$ is at the vertex $$B$$, which is an endpoint of $$BC$$. So $$\angle B$$ touches $$BC$$.
• $$\angle C$$ is at the vertex $$C$$, also an endpoint of $$BC$$. So $$\angle C$$ touches $$BC$$.

Thus the side $$BC$$ is included between the two given angles $$\angle B$$ and $$\angle C$$. The corresponding statement holds in $$\triangle XYZ$$ for side $$YZ$$ and angles $$\angle Y$$, $$\angle Z$$.

This matches the ASA (Angle–Side–Angle) condition: two angles and the side included between them of one triangle equal the corresponding parts of the other.

ASA guarantees congruence. Hence

\[ \triangle ABC \cong \triangle XYZ. \]

No — every triangle with these ASA measurements is congruent to every other. Let us see this from the construction.

Step 1. Draw a segment $$BC$$ of length $$5\,\mathrm{cm}$$ as the base.

Step 2. At $$B$$, use a protractor to draw a ray making $$\angle ABC = 50^{\circ}$$ with $$BC$$ (on the upper side).

Step 3. At $$C$$, similarly draw a ray making $$\angle BCA = 30^{\circ}$$ with $$CB$$ (on the same upper side).

Step 4. The two rays meet at a unique point $$A$$. Then $$\triangle ABC$$ is the required triangle.

Every classmate who follows the same steps ends up with exactly the same triangle (up to rotation or flipping the page) — the two rays from $$B$$ and $$C$$ cross in exactly one point above the base. Hence the ASA data determines a unique triangle: no two non-congruent triangles can have these measurements.

Thus $$\triangle ABC \cong \triangle XYZ$$ is guaranteed, in agreement with the ASA condition.

Since $$O$$ is the midpoint of $$AD$$, we have

$$AO = OD.$$

Since $$O$$ is the midpoint of $$BC$$, we have

$$BO = OC.$$

Also, $$AD$$ and $$BC$$ intersect at $$O$$, so $$\angle AOB$$ and $$\angle DOC$$ are vertically opposite angles, giving

$$\angle AOB = \angle DOC.$$

Now compare $$\triangle AOB$$ and $$\triangle DOC$$. We have:

$$AO = DO,\quad BO = CO,\quad \angle AOB = \angle DOC.$$

The equal angle is included between the two pairs of equal sides. So by the SAS condition,

$$\triangle AOB \cong \triangle DOC.$$

Corresponding sides of congruent triangles are equal, so

\[ AB = DC. \]

Hence the lengths $$AB$$ and $$CD$$ are equal.

From the congruence $$\triangle AOB \cong \triangle DOC$$, all corresponding parts are equal ("CPCT"). We have already used

$$AO = DO,\quad BO = CO,\quad \angle AOB = \angle DOC$$

as inputs. The remaining corresponding parts give us the following extra equalities:

Corresponding sides.

$$AB = DC.$$

Corresponding angles.

$$\angle OAB = \angle ODC \;\;(\text{i.e., } \angle DAB \text{ measured at } A \text{ on the segment } AB) $$

$$\angle OBA = \angle OCD.$$

The equality $$\angle OAB = \angle ODC$$ means that $$\angle A$$ (in $$\triangle AOB$$) equals $$\angle D$$ (in $$\triangle DOC$$). These are alternate angles for the lines $$AB$$ and $$CD$$ cut by the transversal $$AD$$. Since the alternate angles are equal, we further conclude that

\[ AB \parallel CD. \]

So, besides $$AB = CD$$, we also get $$\angle OAB = \angle ODC$$, $$\angle OBA = \angle OCD$$, and the two sides $$AB$$ and $$CD$$ are parallel.

From the figure, the given data for $$\triangle ABC$$ is

$$AB = 7\,\mathrm{cm},\quad BC = 5\,\mathrm{cm},\quad \angle B = 47^{\circ},$$

and for $$\triangle XYZ$$:

$$XZ = 7\,\mathrm{cm},\quad YZ = 5\,\mathrm{cm},\quad \angle Z = 47^{\circ}.$$

In each triangle the given angle is at the vertex where the two given sides meet:

• In $$\triangle ABC$$, the angle $$\angle B$$ is at vertex $$B$$, which is the common endpoint of $$AB$$ and $$BC$$. So $$\angle B$$ is the included angle between $$AB$$ and $$BC$$.
• In $$\triangle XYZ$$, the angle $$\angle Z$$ is the common endpoint of $$XZ$$ and $$YZ$$. So $$\angle Z$$ is the included angle between $$XZ$$ and $$YZ$$.

Matching the equal pairs:

$$AB = XZ = 7\,\mathrm{cm},\quad BC = YZ = 5\,\mathrm{cm},\quad \angle B = \angle Z = 47^{\circ}.$$

By the SAS criterion (two sides and the included angle), the two triangles are congruent. The correspondence of vertices is $$A \leftrightarrow X$$, $$B \leftrightarrow Z$$, $$C \leftrightarrow Y$$. Hence

\[ \triangle ABC \cong \triangle XZY. \]

In the figure, $$AB$$ and $$CD$$ are two parallel segments intersecting at $$O$$ (the two segments $$AC$$ and $$BD$$ cross at $$O$$, splitting the figure into two triangles $$\triangle AOB$$ and $$\triangle COD$$).

Also, $$AB$$ and $$CD$$ are the segments joining opposite ends. Consider the transversals $$AC$$ and $$BD$$ crossing the parallel lines $$AB$$ and $$CD$$.

Alternate angles.

$$\angle OAB = \angle OCD \quad(\text{alternate interior angles for transversal } AC),$$

$$\angle OBA = \angle ODC \quad(\text{alternate interior angles for transversal } BD).$$

Also we are given $$AB = CD$$.

Congruence of the triangles. In $$\triangle OAB$$ and $$\triangle OCD$$,

$$\angle OAB = \angle OCD,\quad AB = CD,\quad \angle OBA = \angle ODC.$$

Here the equal side $$AB = CD$$ is included between the two pairs of equal angles. By the ASA condition,

\[ \triangle OAB \cong \triangle OCD. \]

Other equal parts. By CPCT (corresponding parts of congruent triangles):

$$OA = OC \quad(\text{so } O \text{ is the midpoint of } AC),$$

$$OB = OD \quad(\text{so } O \text{ is the midpoint of } BD),$$

$$\angle AOB = \angle COD \quad(\text{these are, of course, also vertically opposite}).$$

Thus, in addition to $$AB = CD$$ we get $$OA = OC$$, $$OB = OD$$, $$\angle OAB = \angle OCD$$, $$\angle OBA = \angle ODC$$, and $$\angle AOB = \angle COD$$.

In $$\triangle ABC$$ and $$\triangle DBC$$ we are given

$$\angle ABC = \angle DBC,\quad \angle ACB = \angle DCB.$$

Both triangles share the side $$BC$$:

$$BC = BC \quad (\text{common}).$$

Look at how $$BC$$ sits with respect to the two pairs of equal angles. The vertex $$B$$ (an endpoint of $$BC$$) carries the angles $$\angle ABC$$ and $$\angle DBC$$; the vertex $$C$$ (the other endpoint of $$BC$$) carries $$\angle ACB$$ and $$\angle DCB$$. So $$BC$$ is the side included between the two given pairs of equal angles in both triangles.

This is the ASA condition. Hence

\[ \triangle ABC \cong \triangle DBC. \]

Now the third pair of angles must also be equal (corresponding parts of congruent triangles). The third angle of $$\triangle ABC$$ is $$\angle BAC$$ (at vertex $$A$$), and the third angle of $$\triangle DBC$$ is $$\angle BDC$$ (at vertex $$D$$). So

$$\angle BAC = \angle BDC,$$

which is what we wanted to show. Yes, the two triangles are congruent.

The figure shows a quadrilateral with the two diagonals drawn. Consider the two triangles $$\triangle ABC$$ and $$\triangle DCB$$ that share the base $$BC$$.

Angles at $$B$$. The angle $$\angle ABC$$ of $$\triangle ABC$$ is the sum of $$\angle ABD$$ and $$\angle DBC$$:

$$\angle ABC = \angle ABD + \angle DBC.$$

The angle $$\angle DCB$$ of $$\triangle DCB$$ is the sum of $$\angle DCA$$ and $$\angle ACB$$:

$$\angle DCB = \angle DCA + \angle ACB.$$

We are given $$\angle ABD = \angle DCA$$ and $$\angle DBC = \angle ACB$$, so adding gives

$$\angle ABC = \angle DCB.$$

Congruence of $$\triangle ABC$$ and $$\triangle DCB$$. We have

$$\angle ACB = \angle DBC \;(\text{given}),$$

$$BC = CB \;(\text{common side}),$$

$$\angle ABC = \angle DCB \;(\text{shown above}).$$

The equal side $$BC$$ is included between the two pairs of equal angles at $$B$$ and $$C$$. By the ASA condition,

\[ \triangle ABC \cong \triangle DCB. \]

The correspondence is $$A \leftrightarrow D,\; B \leftrightarrow C,\; C \leftrightarrow B$$.

All the equal parts. By CPCT,

• $$AB = DC$$,
• $$AC = DB$$,
• $$BC = CB$$ (the common side),
• $$\angle BAC = \angle CDB$$,
• $$\angle ABC = \angle DCB$$,
• $$\angle ACB = \angle DBC$$.

Consequently the point where the diagonals meet also has extra properties (e.g. the two triangles formed at the intersection are congruent), but the primary equal parts of the figure are the ones listed above.

Look at how the given side $$BC$$ sits with respect to the two given angles.

• $$\angle C$$ is at vertex $$C$$, which is an endpoint of $$BC$$. So $$\angle C$$ touches $$BC$$.
• $$\angle A$$ is at vertex $$A$$, which is not an endpoint of $$BC$$. So $$\angle A$$ is opposite to $$BC$$ — it is a non-included angle.

Hence the data is of the form Angle–Angle–Side (AAS) (two angles and a non-included side).

AAS guarantees congruence: if we know two angles of a triangle, we automatically know the third (since the three angles add to $$180^{\circ}$$), so AAS effectively becomes ASA. Let us make that explicit in the next question.

In brief, yes — $$\triangle ABC \cong \triangle XYZ$$ by the AAS condition.

The sum of the three angles of a triangle is $$180^{\circ}$$.

In $$\triangle ABC$$:

$$\angle A + \angle B + \angle C = 180^{\circ}.$$

Substituting $$\angle A = 35^{\circ}$$ and $$\angle C = 75^{\circ}$$:

$$35^{\circ} + \angle B + 75^{\circ} = 180^{\circ}$$

$$\angle B + 110^{\circ} = 180^{\circ}$$

$$\angle B = 180^{\circ} - 110^{\circ} = 70^{\circ}.$$

Similarly, in $$\triangle XYZ$$:

$$\angle X + \angle Y + \angle Z = 180^{\circ}$$

$$35^{\circ} + \angle Y + 75^{\circ} = 180^{\circ}$$

$$\angle Y = 70^{\circ}.$$

So $$\angle B = \angle Y = 70^{\circ}$$.

Yes. We now have

$$\angle B = \angle Y = 70^{\circ},\quad BC = YZ = 4\,\mathrm{cm},\quad \angle C = \angle Z = 75^{\circ}.$$

The side $$BC$$ (in $$\triangle ABC$$) lies between the two angles $$\angle B$$ and $$\angle C$$; similarly $$YZ$$ lies between $$\angle Y$$ and $$\angle Z$$ in $$\triangle XYZ$$. So the side is included between the two pairs of equal angles.

This is exactly the ASA condition. Hence

\[ \triangle ABC \cong \triangle XYZ. \]

This is how the AAS condition (question 27) reduces to ASA: once we compute the third angle using the $$180^{\circ}$$ rule, the equal side becomes an included side.

Both triangles are right-angled at $$B$$ and $$Y$$ respectively. In each triangle:

• The hypotenuse (the side opposite the right angle) is $$AC$$ in $$\triangle ABC$$ and $$XZ$$ in $$\triangle XYZ$$. We are given $$AC = XZ = 5\,\mathrm{cm}$$.
• One of the legs (a side adjacent to the right angle) is $$BC$$ in $$\triangle ABC$$ and $$YZ$$ in $$\triangle XYZ$$. We are given $$BC = YZ = 4\,\mathrm{cm}$$.
• The right angles are $$\angle B = \angle Y = 90^{\circ}$$.

This matches the RHS (Right angle–Hypotenuse–Side) condition. The RHS condition guarantees congruence of right-angled triangles. Hence

\[ \triangle ABC \cong \triangle XYZ. \]

No — every triangle satisfying these RHS measurements is congruent to every other. Let us verify by constructing the triangle.

Step 1. Draw the base $$QR = 4\,\mathrm{cm}$$.

Step 2. At $$Q$$, using a set-square or protractor, draw a line $$l$$ perpendicular to $$QR$$ (making $$90^{\circ}$$ with $$QR$$).

Step 3. With centre $$R$$ and radius $$5\,\mathrm{cm}$$, draw an arc that cuts the line $$l$$ at a point $$P$$.

Step 4. Join $$PR$$.

Then $$\triangle PQR$$ has $$QR = 4\,\mathrm{cm}$$, $$\angle Q = 90^{\circ}$$ and hypotenuse $$PR = 5\,\mathrm{cm}$$ — the required triangle.

Every classmate who follows the same construction gets a triangle congruent to this one. (See the next question for why the arc gives essentially one triangle.) So there are no non-congruent triangles with the given measurements.

Yes. The line $$l$$ (perpendicular to $$QR$$ at $$Q$$) extends both above and below the base $$QR$$. The arc from $$R$$ of radius $$5\,\mathrm{cm}$$ cuts $$l$$ at two points — one above $$QR$$ (call it $$P$$) and one below $$QR$$ (call it $$P'$$).

So we do get a second triangle $$\triangle P'QR$$ with

$$QR = 4\,\mathrm{cm},\quad \angle Q = 90^{\circ},\quad P'R = 5\,\mathrm{cm}.$$

However, $$\triangle P'QR$$ is just the reflection of $$\triangle PQR$$ across the line $$QR$$ — it is the same triangle turned upside down. In particular, both have

$$QP = QP' \quad\text{and}\quad PR = P'R = 5\,\mathrm{cm}.$$

By the SAS condition (or SSS), $$\triangle PQR \cong \triangle P'QR$$. So even though the arc meets the line at two points, the two resulting triangles are congruent — the size and shape are the same, only the orientation is different.

Hence the RHS data really does determine a unique triangle up to congruence: $$\triangle ABC \cong \triangle XYZ$$.

The important isosceles-triangle fact is: angles opposite equal sides are equal. Let us derive it in our case.

In $$\triangle ABC$$, draw the altitude from $$A$$ to $$BC$$; let its foot be $$D$$. Then $$\angle ADB = \angle ADC = 90^{\circ}$$.

Compare $$\triangle ADB$$ and $$\triangle ADC$$:

$$AB = AC \quad (\text{given, and these are the hypotenuses of the two right triangles}),$$

$$AD = AD \quad (\text{common side}),$$

$$\angle ADB = \angle ADC = 90^{\circ}.$$

So the two right-angled triangles satisfy the RHS condition:

$$\triangle ADB \cong \triangle ADC.$$

By CPCT, $$\angle B = \angle ABD = \angle ACD = \angle C$$.

Hence $$\angle B = \angle C$$: the two base angles of an isosceles triangle are equal.

(To actually compute them from $$\angle A = 80^{\circ}$$, use $$\angle A + \angle B + \angle C = 180^{\circ}$$ — done in the next question.)

We use the two facts

$$\angle B = \angle C \quad(\text{isosceles with } AB = AC),$$

$$\angle A + \angle B + \angle C = 180^{\circ}\quad(\text{angle sum of a triangle}).$$

Substituting $$\angle A = 80^{\circ}$$ and $$\angle C = \angle B$$ into the angle-sum equation:

$$80^{\circ} + \angle B + \angle B = 180^{\circ}$$

$$2\angle B = 180^{\circ} - 80^{\circ} = 100^{\circ}$$

$$\angle B = \dfrac{100^{\circ}}{2} = 50^{\circ}.$$

Hence

\[ \angle B = \angle C = 50^{\circ}. \]

In an equilateral triangle $$\triangle ABC$$, all three sides are equal:

$$AB = BC = CA.$$

Apply the isosceles-triangle result (angles opposite equal sides are equal) to each pair of equal sides.

From $$AB = AC$$: the angles opposite them are $$\angle C$$ (opposite $$AB$$) and $$\angle B$$ (opposite $$AC$$), so

$$\angle B = \angle C.$$

From $$AB = BC$$: the angles opposite them are $$\angle C$$ (opposite $$AB$$) and $$\angle A$$ (opposite $$BC$$), so

$$\angle A = \angle C.$$

Combining: $$\angle A = \angle B = \angle C$$.

So all three angles of an equilateral triangle are equal.

Let each of the three equal angles of an equilateral triangle be $$x$$. The angles of any triangle add up to $$180^{\circ}$$, so

$$x + x + x = 180^{\circ}$$

$$3x = 180^{\circ}$$

$$x = \dfrac{180^{\circ}}{3} = 60^{\circ}.$$

Hence each angle of an equilateral triangle measures

\[ 60^{\circ}. \]

Construct an equilateral triangle with any convenient side, say $$5\,\mathrm{cm}$$, and measure its angles.

Step 1. Draw a segment $$AB = 5\,\mathrm{cm}$$.

Step 2. With centre $$A$$ and radius $$5\,\mathrm{cm}$$, draw an arc above $$AB$$.

Step 3. With centre $$B$$ and radius $$5\,\mathrm{cm}$$, draw another arc that cuts the first arc at a point $$C$$.

Step 4. Join $$AC$$ and $$BC$$. Then $$AB = AC = BC = 5\,\mathrm{cm}$$, so $$\triangle ABC$$ is equilateral.

Now measure each of $$\angle A$$, $$\angle B$$, $$\angle C$$ with a protractor. In each case the reading is $$60^{\circ}$$, confirming that every angle of an equilateral triangle is $$60^{\circ}$$.

Louvre Museum pyramid (Paris). The front face of the pyramid is a large isosceles triangle, and it is divided by struts into many smaller triangular panels. All the triangular glass panels along each row of the pyramid are of the same shape and size — they are equilateral or isosceles triangles that fit together in a repeating pattern. Every such row of small triangles consists of congruent triangles.

Egyptian Pyramid of Giza. The visible faces of the pyramid are congruent isosceles triangles: each of the four faces has the same base (a side of the square base) and the same slant height (the edge from the apex to the mid-point of a base side). So each face is congruent to every other face.

Dome design. The dome is covered by a triangulated network in which the same equilateral (or nearly equilateral) triangle is repeated many times over. All these small triangular panels have the same size and shape — they are congruent.

Rangoli design. A rangoli of this type is made by rotating and reflecting the same basic triangular motif. Each pink and blue triangle around the centre is congruent to the others of its colour; the whole design is built out of just a few shapes of triangles, each repeated many times.

Howrah (Rabindra Setu) bridge. The bridge's steel framework is made of a lattice of triangles arranged in a repeating truss pattern. Along each horizontal row of the truss, the triangles have the same base and the same height, so all these triangles are congruent. This use of many congruent triangles gives the bridge its strength and rigidity.

The order in which vertices are written in a congruence tells us the correspondence directly. Since $$\triangle AIR \cong \triangle FLY$$, the first-letters match, the second-letters match, and the third-letters match:

Corresponding vertices.

$$A \leftrightarrow F,\quad I \leftrightarrow L,\quad R \leftrightarrow Y.$$

Corresponding sides (formed by consecutive matched vertices).

$$AI \leftrightarrow FL,\quad IR \leftrightarrow LY,\quad AR \leftrightarrow FY.$$

Being corresponding sides of congruent triangles, they are equal in length: $$AI = FL$$, $$IR = LY$$, $$AR = FY$$.

Corresponding angles (angles at matched vertices).

$$\angle A \leftrightarrow \angle F,\quad \angle I \leftrightarrow \angle L,\quad \angle R \leftrightarrow \angle Y.$$

Being corresponding angles of congruent triangles, they are equal: $$\angle A = \angle F$$, $$\angle I = \angle L$$, $$\angle R = \angle Y$$.

The segments $$AD$$ and $$BC$$ cross at the point $$O$$, forming two triangles $$\triangle AOB$$ and $$\triangle DOC$$ on opposite sides of $$O$$.

Compare $$\triangle AOB$$ and $$\triangle DOC$$:

$$OA = OD \quad (\text{given}),$$

$$OB = OC \quad (\text{given}),$$

$$\angle AOB = \angle DOC \quad (\text{vertically opposite angles at } O).$$

Two sides and the included angle of $$\triangle AOB$$ are equal to the corresponding parts of $$\triangle DOC$$. By the SAS criterion,

$$\triangle AOB \cong \triangle DOC.$$

By CPCT (corresponding parts of congruent triangles),

$$\angle OAB = \angle ODC,$$

i.e. $$\angle BAD = \angle CDA$$ (using the fact that $$A$$, $$O$$, $$D$$ are collinear on the transversal $$AD$$).

Now, $$AD$$ is a transversal cutting the two lines $$AB$$ and $$CD$$, and $$\angle BAD$$ and $$\angle CDA$$ are alternate interior angles. Since these alternate angles are equal, the two lines are parallel:

\[ AB \parallel CD. \]

Part 1: $$\triangle ABC \cong \triangle ADC$$.

In the square $$ABCD$$, all four sides are equal and every angle is $$90^{\circ}$$. The diagonal $$AC$$ splits the square into $$\triangle ABC$$ and $$\triangle ADC$$.

Compare $$\triangle ABC$$ and $$\triangle ADC$$:

$$AB = AD \quad (\text{sides of the square}),$$

$$BC = DC \quad (\text{sides of the square}),$$

$$AC = AC \quad (\text{common diagonal}).$$

By the SSS criterion, $$\triangle ABC \cong \triangle ADC$$.

Part 2: Is $$\triangle ABC$$ also congruent to $$\triangle CDA$$?

Check the correspondence given by writing $$\triangle ABC \cong \triangle CDA$$: $$A \leftrightarrow C$$, $$B \leftrightarrow D$$, $$C \leftrightarrow A$$. Under this correspondence:

$$AB \text{ pairs with } CD, \quad BC \text{ pairs with } DA, \quad CA \text{ pairs with } AC.$$

Since $$AB = CD$$, $$BC = DA$$ (sides of the square) and $$CA = AC$$, by SSS the congruence holds too:

$$\triangle ABC \cong \triangle CDA.$$

So $$\triangle ABC$$ is congruent to $$\triangle ADC$$ in two different ways — as $$\triangle ABC \cong \triangle ADC$$ and as $$\triangle ABC \cong \triangle CDA$$. (Geometrically: the diagonal $$AC$$ is an axis of symmetry of the square, so we can either flip $$\triangle ABC$$ across $$AC$$ to get $$\triangle ADC$$ in the first way, or additionally rotate through $$180^{\circ}$$ to get the second correspondence.)

Part 3: More examples of "two ways".

Take an isosceles triangle $$\triangle PQR$$ with $$PQ = PR$$. Compared with itself, we have

• $$\triangle PQR \cong \triangle PQR$$ (the identity correspondence), and
• $$\triangle PQR \cong \triangle PRQ$$ (flip across the axis of symmetry through $$P$$; uses $$PQ = PR$$, common side $$P$$-something, and equal base angles).

Another example: two congruent isosceles trapezium halves. In general, any triangle that has a line of symmetry — an isosceles triangle — is congruent to itself in exactly two ways.

Part 4: Six different ways.

A triangle can be re-labelled in $$3! = 6$$ orders. If every such re-labelling gives a valid congruence, we get six different ways.

This happens precisely when all three sides (and all three angles) of the triangle are equal, i.e. when the triangle is equilateral. For an equilateral triangle $$\triangle ABC$$ (with $$AB = BC = CA$$), we have

$$\triangle ABC \cong \triangle ABC,\; \triangle ACB,\; \triangle BCA,\; \triangle BAC,\; \triangle CAB,\; \triangle CBA,$$

all six of which are correct. So any equilateral triangle is congruent to itself in six different ways.

From the figure, $$B$$ and $$C$$ are two points on the circle, and $$A$$ is its centre. So $$AB$$ and $$AC$$ are both radii of the circle and are therefore equal:

$$AB = AC.$$

Hence $$\triangle ABC$$ is isosceles, with the two equal sides meeting at the vertex $$A$$. The angles opposite the equal sides are equal:

$$\angle B = \angle C.$$

The given angle at $$A$$ is $$\angle BAC = 120^{\circ}$$. Using the angle sum of the triangle:

$$\angle A + \angle B + \angle C = 180^{\circ}$$

$$120^{\circ} + \angle B + \angle B = 180^{\circ}$$

$$2\angle B = 180^{\circ} - 120^{\circ} = 60^{\circ}$$

$$\angle B = \dfrac{60^{\circ}}{2} = 30^{\circ}.$$

Hence

\[ \angle B = \angle C = 30^{\circ}. \]

The figure is a rectangle $$ACBD$$-like frame in which many segments have been drawn, marking off several triangles. Some sides carry the same number of tick-marks, indicating equal length. To find the missing angles we use two facts throughout:

• F1 (angle sum): The three angles of a triangle add up to $$180^{\circ}$$.
• F2 (isosceles triangle): In a triangle, the angles opposite equal sides are equal.

We apply these facts to each small triangle in the figure.

Step-by-step method for each triangle.

Step (i). Look at the two tick-marked sides of the triangle. If they carry the same marking, the two angles opposite them are equal (Fact F2). This gives you a pair of equal unknown angles.

Step (ii). Add up the three angles and set the sum equal to $$180^{\circ}$$ (Fact F1). This gives you one linear equation in the unknowns, which can be solved.

Step (iii). If a corner of the frame carries a right-angle box, use $$90^{\circ}$$ there. If a straight line passes through an interior point, use the fact that angles on a straight line add to $$180^{\circ}$$ to split or combine adjacent angles.

Two illustrative calculations from the figure.

1. Consider the small triangle at the top-left with vertices $$C$$, $$R$$ (on $$CD$$), $$U$$ (on $$CA$$). We have $$\angle C = 90^{\circ}$$ (rectangle corner) and $$\angle CRU = 34^{\circ}$$. By F1, the third angle is

$$\angle CUR = 180^{\circ} - 90^{\circ} - 34^{\circ} = 56^{\circ}.$$

2. Consider a triangle in which the two equal tick-marks show two sides equal, and one angle is $$68^{\circ}$$. By F2 the other two angles are equal, and by F1

$$2\theta + 68^{\circ} = 180^{\circ}\;\;\Rightarrow\;\;\theta = \dfrac{180^{\circ} - 68^{\circ}}{2} = 56^{\circ}.$$

So each equal base angle is $$56^{\circ}$$.

Working around the figure in this way, using each triangle in turn, every missing angle can be filled in. The three main tools are the same everywhere: the angle-sum $$180^{\circ}$$, the isosceles property (equal tick marks $$\Rightarrow$$ equal base angles), and the right angles at the corners of the rectangle.

Some of the missing angles that come out of this process are $$56^{\circ}$$ (as computed above), $$78^{\circ} = 180^{\circ} - 34^{\circ} - 68^{\circ}$$ inside triangles with sides $$34^{\circ}, 68^{\circ}$$ marked at the top, $$60^{\circ}$$ in equilateral-looking triangles (three sides marked equal), and so on. Every missing angle is found by one of these two rules.

The figure is a $$5 \times 5$$ grid of unit squares in which the central square is coloured green. So the white region consists of $$25 - 1 = 24$$ unit squares.

Splitting these $$24$$ white squares into $$6$$ smaller congruent regions means each region must contain

$$\dfrac{24}{6} = 4 \;\text{unit squares},$$

and the six pieces must all have the same shape and size.

One neat way to do this uses rotational symmetry about the centre. The whole white region is symmetric under a $$90^{\circ}$$ rotation about the centre; if we further use reflections, it has all the symmetries of a square. A convenient choice is to split the white region into six identical L-shaped tetrominoes (each an L-shape covering 4 squares).

An explicit description (using row–column coordinates $$(r,c)$$ with rows numbered $$1$$ to $$5$$ from top and columns $$1$$ to $$5$$ from left; the green square is $$(3,3)$$):

1. Top-left L: $$(1,1), (1,2), (2,1), (2,2)$$.
2. Top-right L: $$(1,4), (1,5), (2,4), (2,5)$$.
3. Bottom-left L: $$(4,1), (4,2), (5,1), (5,2)$$.
4. Bottom-right L: $$(4,4), (4,5), (5,4), (5,5)$$.
5. Top strip: $$(1,3), (2,3), (3,1), (3,2)$$ — arranged as an L wrapping around the green square from the top-left.
6. Bottom strip: $$(3,4), (3,5), (4,3), (5,3)$$ — the L wrapping around the green square from the bottom-right.

Each of these six pieces contains exactly $$4$$ unit squares and has the same L (or $$2\times2$$-square) shape, so all six are congruent.

(Any other partition into six congruent tetrominoes obtained by rotating this pattern by $$90^{\circ}$$, $$180^{\circ}$$, etc. is also a correct answer.)