Chapter 4: Quadrilaterals

A quadrilateral is a closed plane figure bounded by exactly four straight line segments such that:

• each vertex is shared by exactly two sides,
• no two sides cross each other, and
• the figure is closed (no gaps or free ends).

Figures (i), (ii) and (iii) satisfy all three conditions — they are made of four straight sides joined end-to-end, forming a closed shape.

The other figures fail one of these conditions. For example, a figure could:

• have curved sides (not made only of straight line segments),
• be open (the sides do not close up), or
• have more than four sides, or two sides that cross each other.

Because of any such failure, they cannot be called quadrilaterals.

Yes. The usual definition — a quadrilateral with opposite sides equal and all four angles equal to $$90^{\circ}$$ — contains more information than is actually needed. A rectangle can be defined equivalently in several shorter ways, each of which forces every other rectangle property to hold. Some standard equivalent definitions are:

1. Diagonal definition: a quadrilateral whose diagonals are equal in length and bisect each other.
2. Angle definition: a quadrilateral in which all four angles equal $$90^{\circ}$$.
3. Parallelogram definition: a parallelogram in which one angle is $$90^{\circ}$$ (this forces all four angles to be $$90^{\circ}$$).
4. Parallelogram definition (via diagonals): a parallelogram whose diagonals are equal in length.

Each of these gives exactly the same shape as the usual definition, so any one of them can be taken as the definition of a rectangle.

No. To use the SAS congruence rule, the angle used must be the angle included between the two given equal sides of the triangle.

In $$\triangle AOD$$, the side $$AO$$ and the side $$AD$$ do not form the angle $$\angle AOB$$. The angle between $$AO$$ and $$OD$$ in $$\triangle AOD$$ is $$\angle AOD$$, not $$\angle AOB$$. Similarly, in $$\triangle COB$$, the angle between $$CO$$ and $$OB$$ is $$\angle COB$$, not $$\angle COD$$.

Also, the angle $$\angle AOB$$ is not even an angle of $$\triangle AOD$$, and $$\angle COD$$ is not an angle of $$\triangle COB$$. So this set of equalities does not match the SAS pattern, and the given data cannot be used to conclude $$\triangle AOD \cong \triangle COB$$.

The correct set of equalities is $$AO = CO$$, $$OD = OB$$, and the included angle $$\angle AOD = \angle COB$$ (vertically opposite angles). With these, $$\triangle AOD \cong \triangle COB$$ by SAS.

Let the diagonals $$AC$$ and $$BD$$ of the quadrilateral $$ABCD$$ meet at $$O$$. We are given $$AC = BD$$, $$OA = OC$$, $$OB = OD$$, and one of the angles at $$O$$ is $$60^{\circ}$$.

The four angles at $$O$$ come in two pairs of vertically-opposite angles, and each pair of adjacent angles is a linear pair (adds to $$180^{\circ}$$).

If $$\angle AOB = 60^{\circ}$$, then:

$$\angle COD = \angle AOB = 60^{\circ}$$ (vertically opposite).

$$\angle BOC = 180^{\circ} - 60^{\circ} = 120^{\circ}$$ (linear pair with $$\angle AOB$$).

$$\angle AOD = \angle BOC = 120^{\circ}$$ (vertically opposite).

In $$\triangle AOB$$, $$OA = OB$$, so it is an isosceles triangle. The angles opposite the equal sides are equal, so $$\angle OAB = \angle OBA = a$$.

The angle at $$O$$ inside this triangle is $$\angle AOB = 60^{\circ}$$ (from the previous step).

By the angle-sum property of a triangle,

$$a + a + 60^{\circ} = 180^{\circ}$$

$$2a = 120^{\circ}$$

$$a = 60^{\circ}$$

Applying the same argument to each of the four triangles formed by the diagonals ($$\triangle AOB, \triangle BOC, \triangle COD, \triangle DOA$$), we find that all four are isosceles (since $$OA = OB = OC = OD$$, as the two equal diagonals bisect each other).

Going through each triangle, the base angles come out to be $$60^{\circ}$$, $$30^{\circ}$$, $$60^{\circ}$$, $$30^{\circ}$$ around the quadrilateral. The angle at vertex $$A$$ is $$\angle DAO + \angle OAB = 30^{\circ} + 60^{\circ} = 90^{\circ}$$. Similarly each of the four vertex angles of $$ABCD$$ equals $$90^{\circ}$$.

Also, using the isosceles triangles, we get $$AB = CD$$ and $$AD = BC$$ (opposite sides equal).

Hence $$ABCD$$ is a rectangle.

Since the diagonals of $$ABCD$$ bisect each other, we have $$OA = OC$$ and $$OB = OD$$. Also, vertically opposite angles at $$O$$ are equal.

By the SAS congruence rule, $$\triangle AOB \cong \triangle COD$$, giving $$AB = CD$$. Similarly, $$\triangle BOC \cong \triangle DOA$$, giving $$BC = DA$$.

Hence, the opposite sides of $$ABCD$$ are equal:

$$AB = CD \quad \text{and} \quad BC = DA.$$

Yes. Suppose the two diagonals are equal and bisect each other, and one of the angles between them is $$x$$ (where $$0^{\circ} < x < 180^{\circ}$$). We will show that $$ABCD$$ is always a rectangle, no matter what $$x$$ is.

Since the diagonals bisect each other, $$OA = OC$$ and $$OB = OD$$. Because the diagonals are equal, $$OA = OB = OC = OD$$ (each equals half of the common diagonal length). So each of the four triangles $$\triangle AOB, \triangle BOC, \triangle COD, \triangle DOA$$ is isosceles.

We will show in the next few steps that each angle of $$ABCD$$ turns out to be $$90^{\circ}$$ regardless of $$x$$. Hence $$ABCD$$ is a rectangle for every $$x$$.

The two vertically opposite angles at $$O$$ that equal $$x$$ are $$\angle AOB$$ and $$\angle COD$$. The other two, $$\angle BOC$$ and $$\angle DOA$$, are the linear-pair supplements, so each equals $$180^{\circ} - x$$.

Now look at $$\triangle AOB$$. It is isosceles with $$OA = OB$$, so the base angles $$\angle OAB$$ and $$\angle OBA$$ are equal. Using the angle sum:

$$\angle OAB + \angle OBA + x = 180^{\circ}$$

$$\angle OAB = \angle OBA = \dfrac{180^{\circ} - x}{2} = 90^{\circ} - \dfrac{x}{2}.$$

Similarly, in $$\triangle COD$$ (isosceles with $$OC = OD$$ and included angle $$x$$),

$$\angle OCD = \angle ODC = 90^{\circ} - \dfrac{x}{2}.$$

In $$\triangle BOC$$ (isosceles with $$OB = OC$$ and included angle $$180^{\circ}-x$$),

$$\angle OBC = \angle OCB = \dfrac{180^{\circ} - (180^{\circ}-x)}{2} = \dfrac{x}{2}.$$

Similarly, in $$\triangle DOA$$,

$$\angle ODA = \angle OAD = \dfrac{x}{2}.$$

In $$\triangle AOB$$, $$OA = OB$$, so the base angles $$\angle OAB$$ and $$\angle OBA$$ are equal — call each of them $$a$$. The included angle at the vertex $$O$$ is $$\angle AOB = x$$.

By the angle-sum property,

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

$$2a = 180^{\circ} - x$$

$$a = 90^{\circ} - \dfrac{x}{2}.$$

Consider $$\triangle AOB$$ and $$\triangle COD$$. We have

$$OA = OC, \quad OB = OD, \quad \angle AOB = \angle COD \text{ (vertically opposite)}.$$

By SAS congruence, $$\triangle AOB \cong \triangle COD$$, so the corresponding sides satisfy

$$AB = CD.$$

Similarly, $$\triangle BOC \cong \triangle DOA$$ (by SAS with $$OB = OD$$, $$OC = OA$$ and $$\angle BOC = \angle DOA$$), so

$$BC = DA.$$

Hence, in $$ABCD$$, the opposite sides are equal.

No — this shorter definition is not wrong. As we will show (in Deduction 4), once we insist that all four angles of a quadrilateral are $$90^{\circ}$$, the property that opposite sides are equal follows automatically. So condition (a) is a consequence of condition (b), and stating (b) alone is enough to characterise a rectangle.

No, such a quadrilateral cannot be constructed. Every attempt to draw a quadrilateral with all four angles equal to $$90^{\circ}$$ ends up with opposite sides equal. This suggests that the property “all angles $$= 90^{\circ}$$” is enough to force the shape to be a rectangle.

The reason is proved in the next question (Deduction 4): using the diagonal $$BD$$, we can show $$\triangle BAD \cong \triangle DCB$$, which forces $$AB = DC$$ and $$AD = BC$$.

Look at $$\triangle BAD$$ and $$\triangle DCB$$. Two things are directly visible:

1. $$\angle A = \angle C = 90^{\circ}$$ (given).

2. $$BD$$ is common to both triangles, so $$BD = DB$$.

Let $$\angle 1 = \angle ABD$$ (in $$\triangle BAD$$) and $$\angle 2 = \angle CDB$$ (in $$\triangle DCB$$).

In $$\triangle BAD$$, by the angle-sum property,

$$\angle A + \angle ABD + \angle ADB = 180^{\circ}$$

$$90^{\circ} + \angle 1 + \angle ADB = 180^{\circ}$$

$$\angle 1 = 90^{\circ} - \angle ADB.$$

Now the full angle $$\angle ADC = 90^{\circ}$$, so $$\angle ADB + \angle BDC = 90^{\circ}$$, which gives $$\angle BDC = 90^{\circ} - \angle ADB$$. That is, $$\angle 2 = \angle BDC = 90^{\circ} - \angle ADB = \angle 1$$.

Hence $$\angle 1 = \angle 2$$.

Now in $$\triangle BAD$$ and $$\triangle DCB$$ we have:

• $$\angle A = \angle C = 90^{\circ}$$,
• $$BD = DB$$ (common),
• $$\angle 1 = \angle 2$$.

By ASA (angle–side–angle), $$\triangle BAD \cong \triangle DCB$$. Corresponding sides give $$AB = CD$$ and $$AD = CB$$.

Yes, it is wrong. In a congruence statement, the order of the vertices matters — the vertices in corresponding positions must actually correspond to each other.

From the earlier work, the correct correspondence is

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

which is exactly what $$\triangle BAD \cong \triangle DCB$$ says.

Writing $$\triangle BAD \cong \triangle CDB$$ would mean $$B \leftrightarrow C, A \leftrightarrow D, D \leftrightarrow B$$, i.e. $$BA = CD, AD = DB, BD = CB$$. This is not the correspondence given by the congruence proof, so this statement is wrong.

Yes. In rectangle $$ABCD$$, take $$AD$$ as the transversal cutting $$AB$$ and $$DC$$. The two co-interior (same-side interior) angles it makes are $$\angle A$$ and $$\angle D$$. Since each is $$90^{\circ}$$,

$$\angle A + \angle D = 90^{\circ} + 90^{\circ} = 180^{\circ}.$$

When co-interior angles formed by a transversal add to $$180^{\circ}$$, the two lines it cuts are parallel. Hence,

$$AB \parallel DC.$$

The same argument with $$AB$$ as transversal to $$AD$$ and $$BC$$ gives $$AD \parallel BC$$. So both pairs of opposite sides of a rectangle are parallel.

A square is a special rectangle, so we still need:

• the two strips (diagonals) to be of equal length, and
• joined at their midpoints (diagonals bisect each other).

Both of these together guarantee we get a rectangle. To make it a square, we additionally need all four sides of the resulting quadrilateral to be equal. As we shall see, this happens exactly when the two diagonals are perpendicular to each other. So, apart from being equal and bisecting each other, the strips must also be placed at right angles.

Yes. Recall from Question 10 that if $$x$$ is the angle between the diagonals, the four isosceles triangles $$\triangle AOB, \triangle BOC, \triangle COD, \triangle DOA$$ (with $$OA = OB = OC = OD = \tfrac{d}{2}$$ where $$d$$ is the common diagonal length) have vertex angles $$x, 180^{\circ}-x, x, 180^{\circ}-x$$ at $$O$$.

Using the isosceles-triangle formula for the base (with equal sides $$r = \tfrac{d}{2}$$ and vertex angle $$\theta$$), the side of the quadrilateral opposite the vertex angle $$\theta$$ depends on $$\theta$$. So $$AB$$ and $$CD$$ (from triangles with vertex angle $$x$$) will be equal to each other, and $$BC$$ and $$DA$$ (from vertex angle $$180^{\circ}-x$$) will be equal to each other, but $$AB$$ will equal $$BC$$ only when $$x = 180^{\circ}-x$$, i.e. $$x = 90^{\circ}$$.

Hence, all four sides are equal exactly when the two diagonals meet at $$90^{\circ}$$.

Since $$\triangle BOA \cong \triangle BOC$$, the corresponding angles are equal:

$$\angle BOA = \angle BOC.$$

But $$\angle BOA$$ and $$\angle BOC$$ together form the straight angle $$\angle AOC$$ along the diagonal $$AC$$:

$$\angle BOA + \angle BOC = 180^{\circ}.$$

So $$2 \angle BOA = 180^{\circ}$$, giving

$$\angle BOA = \angle BOC = 90^{\circ}.$$

Hence the two diagonals of a square meet at right angles.

Steps of construction:

1. Draw a line segment $$AC$$ of length $$8 \, \mathrm{cm}$$. This will be one diagonal.
2. Locate its midpoint $$O$$, so that $$OA = OC = 4 \, \mathrm{cm}$$.
3. At $$O$$, using a protractor (or a compass construction), draw a line perpendicular to $$AC$$.
4. On this perpendicular, mark two points $$B$$ and $$D$$ (one on each side of $$AC$$) such that $$OB = OD = 4 \, \mathrm{cm}$$. (This makes $$BD = 8 \, \mathrm{cm}$$, equal to $$AC$$.)
5. Join $$AB, BC, CD, DA$$.

Then $$ABCD$$ is a square because its diagonals $$AC$$ and $$BD$$ are equal ($$8 \, \mathrm{cm}$$), bisect each other at $$O$$, and are perpendicular. On measuring, each side of the square is about $$5.66 \, \mathrm{cm}$$ (which matches $$\dfrac{8}{\sqrt{2}}$$).

A square is obtained by adding one extra condition (all sides equal, or equivalently, diagonals perpendicular) to a rectangle. So every property that Deductions 1 and 2 prove for a rectangle continues to hold for a square:

• Deduction 1 (diagonals equal): the diagonals of a square are equal in length. This is clearly true — indeed, in the construction $$AC = BD = 8 \, \mathrm{cm}$$.
• Deduction 2 (diagonals bisect each other): in a square the two diagonals bisect each other. Again true — $$O$$ is the midpoint of both diagonals.
• All four angles of a square are $$90^{\circ}$$ (as in a rectangle).
• Opposite sides of a square are equal and parallel (as in a rectangle) — in fact, all four sides are equal.

So the reasoning of Deduction 1 and Deduction 2 applies word-for-word to a square, confirming that a square inherits every rectangle property.

Consider the diagonal $$AC$$ of the square $$ABCD$$; it splits the square into two right-angled isosceles triangles.

In $$\triangle ADC$$: angle at $$D$$ is $$90^{\circ}$$, and the two sides $$AD = DC$$ give equal base angles, so $$\angle 1 = \angle DAC = \angle 3 = \angle DCA = 45^{\circ}$$.

Similarly, in $$\triangle ABC$$: angle at $$B$$ is $$90^{\circ}$$ and $$AB = BC$$, so $$\angle 2 = \angle BAC$$ and $$\angle 4 = \angle BCA$$ are equal. Using the angle sum,

$$\angle 2 + \angle 4 + 90^{\circ} = 180^{\circ},$$

and since $$\angle 2 = \angle 4$$, each equals $$45^{\circ}$$.

Hence $$\angle 1 = \angle 2 = \angle 3 = \angle 4 = 45^{\circ}$$. This shows that in a square each diagonal bisects the corresponding pair of vertex angles into two $$45^{\circ}$$ parts.

No, it is not possible. Whatever three $$90^{\circ}$$ angles we mark, the fourth angle is forced to be $$90^{\circ}$$ too, as we prove in the next step using the angle-sum property of a quadrilateral.

The sum of the four interior angles of any quadrilateral is $$360^{\circ}$$. If three of the angles are each $$90^{\circ}$$, then the fourth angle equals

$$360^{\circ} - (90^{\circ} + 90^{\circ} + 90^{\circ}) = 360^{\circ} - 270^{\circ} = 90^{\circ}.$$

So the fourth angle is forced to be $$90^{\circ}$$ — it cannot be anything else. Hence no such quadrilateral exists.

Yes. A quadrilateral whose both pairs of opposite sides are parallel is called a parallelogram. Every rectangle is a parallelogram, but a general parallelogram need not have all angles $$90^{\circ}$$ — its adjacent angles need only be supplementary.

For example, a parallelogram with adjacent sides $$4 \, \mathrm{cm}$$ and $$5 \, \mathrm{cm}$$ and an included angle of $$60^{\circ}$$ has all four opposite sides parallel, but is not a rectangle because its angles are not $$90^{\circ}$$. A rhombus that is not a square is another example.

Steps of construction for a parallelogram $$ABCD$$ with $$AB = 5 \, \mathrm{cm}$$, $$AD = 4 \, \mathrm{cm}$$ and $$\angle A = 60^{\circ}$$:

1. Draw $$AB = 5 \, \mathrm{cm}$$ using the ruler.
2. At $$A$$, using a protractor (or a $$60^{\circ}$$ set-square), draw a ray making an angle of $$60^{\circ}$$ with $$AB$$. On this ray mark $$D$$ so that $$AD = 4 \, \mathrm{cm}$$.
3. Through $$D$$, draw a line parallel to $$AB$$ (use a ruler and a set-square: slide the set-square along the ruler placed along $$AB$$ so that a fixed edge of the set-square passes through $$D$$; the second parallel edge gives the parallel line).
4. Through $$B$$, similarly draw a line parallel to $$AD$$.
5. These two lines meet at $$C$$. Then $$ABCD$$ is the required parallelogram (opposite sides parallel, but $$\angle A = 60^{\circ} \ne 90^{\circ}$$, so it is not a rectangle).

Yes. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel. In Question 17 we showed that in a rectangle $$ABCD$$ (with all angles $$90^{\circ}$$),

$$AB \parallel DC \quad \text{and} \quad AD \parallel BC$$

because the co-interior angles on each pair add up to $$180^{\circ}$$. Hence every rectangle has both pairs of opposite sides parallel, i.e. every rectangle is a parallelogram.

Steps of construction for parallelogram $$ABCD$$ with $$AB = 5 \, \mathrm{cm}, AD = 4 \, \mathrm{cm}$$ and $$\angle A = 30^{\circ}$$:

1. Draw $$AB = 5 \, \mathrm{cm}$$.
2. At $$A$$ construct $$\angle BAX = 30^{\circ}$$ using a protractor. On the ray $$AX$$, mark $$D$$ with $$AD = 4 \, \mathrm{cm}$$.
3. With centre $$B$$ and radius $$4 \, \mathrm{cm}$$, draw an arc.
4. With centre $$D$$ and radius $$5 \, \mathrm{cm}$$, draw another arc, cutting the first arc at $$C$$.
5. Join $$BC$$ and $$DC$$. Then $$ABCD$$ is the required parallelogram (with $$BC = AD = 4 \, \mathrm{cm}$$ and $$DC = AB = 5 \, \mathrm{cm}$$).

Take the parallelogram $$ABCD$$ of Question 29 with $$\angle A = 30^{\circ}$$ and adjacent sides $$AB = 5 \, \mathrm{cm}, AD = 4 \, \mathrm{cm}$$.

Sides. In a parallelogram, opposite sides are equal, so

$$DC = AB = 5 \, \mathrm{cm}, \qquad BC = AD = 4 \, \mathrm{cm}.$$

Angles. $$AD \parallel BC$$ with $$AB$$ as transversal, so co-interior angles are supplementary:

$$\angle A + \angle B = 180^{\circ} \implies \angle B = 180^{\circ} - 30^{\circ} = 150^{\circ}.$$

Similarly, $$AB \parallel DC$$ with $$BC$$ as transversal gives $$\angle B + \angle C = 180^{\circ}$$, so $$\angle C = 30^{\circ}$$. And $$\angle C + \angle D = 180^{\circ}$$ gives $$\angle D = 150^{\circ}$$.

Yes. In every parallelogram, opposite angles are equal.

Reason. Let $$ABCD$$ be a parallelogram, so $$AB \parallel DC$$ and $$AD \parallel BC$$. Taking $$AD$$ as transversal to the parallel lines $$AB$$ and $$DC$$,

$$\angle A + \angle D = 180^{\circ} \quad \text{(co-interior)}.$$

Taking $$DC$$ as transversal to the parallel lines $$AD$$ and $$BC$$,

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

Subtracting, $$\angle A = \angle C$$. In the same way, $$\angle B = \angle D$$.

Yes. The correct congruence in parallelogram $$ABCD$$ (with diagonal $$BD$$) is $$\triangle ABD \cong \triangle CDB$$, coming from

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

The correspondence given by this is $$A\leftrightarrow C, B\leftrightarrow D, D\leftrightarrow B$$.

Writing $$\triangle ABD \cong \triangle CBD$$ would demand the correspondence $$A\leftrightarrow C, B\leftrightarrow B, D\leftrightarrow D$$, i.e. $$AB=CB, BD=BD, AD=CD$$. This is not what the given equalities say. Hence the statement $$\triangle ABD \cong \triangle CBD$$ is wrong.

No. In general the diagonals of a parallelogram are not equal.

On measuring the diagonals of the parallelogram from Question 29 ($$\angle A = 30^{\circ}$$, $$AB = 5 \, \mathrm{cm}$$, $$AD = 4 \, \mathrm{cm}$$), one finds $$AC \approx 8.72 \, \mathrm{cm}$$ and $$BD \approx 2.68 \, \mathrm{cm}$$ — very different values. So the diagonals are unequal in a general parallelogram; they turn out to be equal only in the special case of a rectangle (and hence a square).

Yes — the diagonals of every parallelogram bisect each other.

Proof. Let the diagonals $$AC$$ and $$BD$$ of the parallelogram $$ABCD$$ meet at $$O$$. Consider $$\triangle AOB$$ and $$\triangle COD$$:

• $$AB = CD$$ (opposite sides of a parallelogram),
• $$\angle OAB = \angle OCD$$ (alternate angles, since $$AB \parallel DC$$ with $$AC$$ as transversal),
• $$\angle OBA = \angle ODC$$ (alternate angles, since $$AB \parallel DC$$ with $$BD$$ as transversal).

By ASA, $$\triangle AOB \cong \triangle COD$$. Corresponding sides give $$OA = OC$$ and $$OB = OD$$. Hence $$O$$ is the midpoint of both $$AC$$ and $$BD$$.

Yes, it is wrong. In the parallelogram $$EASY$$ (with diagonals $$ES$$ and $$AY$$ intersecting at $$O$$), the two triangles formed on opposite sides of $$O$$ are $$\triangle AOE$$ and $$\triangle YOS$$ (or equivalently $$\triangle SOY$$ traversed the other way).

Using the standard argument, we have $$AE = SY$$ (opposite sides), $$\angle EAO = \angle YSO$$ (alternate angles), and $$\angle AEO = \angle SYO$$ (alternate angles). This gives the correspondence

$$A \leftrightarrow S, \quad O \leftrightarrow O, \quad E \leftrightarrow Y,$$

so the correct statement is $$\triangle AOE \cong \triangle SOY$$ only if vertices are matched exactly like this. If someone writes it as $$\triangle AOE \cong \triangle SOY$$ but really means $$A\leftrightarrow Y, O\leftrightarrow O, E\leftrightarrow S$$, the congruence statement is wrong because the vertex order does not reflect the actual matching of corresponding sides.

In short: a congruence statement is wrong whenever the vertex order does not put corresponding vertices in the same position on both sides.

No, the diagonals of a general parallelogram meet at no fixed angle — the angle depends on the shape of the parallelogram. In the parallelogram of Question 29, for example, the diagonals meet at about $$14^{\circ}$$ and $$166^{\circ}$$. A different parallelogram (different side lengths or vertex angle) gives a different intersection angle.

The diagonals meet at a fixed angle only in special cases — for instance, they are perpendicular ($$90^{\circ}$$) precisely when the parallelogram is a rhombus (or a square).

No — squares are not the only quadrilaterals with all sides equal. To see this, start the construction:

1. Draw a segment $$AB$$ of some convenient length, say $$4 \, \mathrm{cm}$$.
2. At $$A$$, draw another segment $$AD$$ of the same length $$4 \, \mathrm{cm}$$, making an angle $$\angle DAB$$ that is not $$90^{\circ}$$ (say $$60^{\circ}$$).

So far $$AD = AB$$, but $$\angle DAB \ne 90^{\circ}$$, so if we manage to complete this into a quadrilateral with all sides equal, it will be a rhombus that is not a square.

Yes. Continue the construction of Question 37 as follows.

1. Let $$s$$ be the common side length ($$AD = AB = s$$).
2. With centre $$B$$ and radius $$s$$, draw an arc.
3. With centre $$D$$ and radius $$s$$, draw another arc.
4. These arcs meet at a point $$C$$ (on the far side of $$BD$$ from $$A$$). Join $$BC$$ and $$DC$$.

Since $$BC = s$$ and $$DC = s$$ by construction, all four sides $$AB, BC, CD, DA$$ are equal to $$s$$. Hence $$ABCD$$ is a quadrilateral with all sides equal. This is a rhombus. Because we chose $$\angle DAB \ne 90^{\circ}$$, it is a rhombus that is not a square.

Since all four sides of a rhombus are equal, opposite sides are also equal. In particular, $$AB \parallel DC$$ and $$AD \parallel BC$$ (a rhombus is a parallelogram — see Question 40 for the reasoning). Hence the rules for parallelogram angles apply: adjacent angles are supplementary and opposite angles are equal.

If we constructed the rhombus with $$\angle A = 60^{\circ}$$, then:

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

$$\angle C = \angle A = 60^{\circ},$$

$$\angle D = \angle B = 120^{\circ}.$$

In the rhombus $$GAME$$, join the diagonal $$AE$$. In $$\triangle GAE$$ and $$\triangle MAE$$:

• $$GA = MA$$ (sides of the rhombus),
• $$GE = ME$$ (sides of the rhombus),
• $$AE = AE$$ (common side).

By the SSS congruence rule, $$\triangle GAE \cong \triangle MAE$$.

Corresponding angles then give $$\angle GAE = \angle MAE$$ and $$\angle GEA = \angle MEA$$ — i.e. the diagonal $$AE$$ bisects both $$\angle A$$ and $$\angle E$$ of the rhombus.

We draw one big oval labelled “Parallelograms”. Inside it, we draw two smaller ovals — one labelled “Rhombuses” and one labelled “Rectangles”. Both smaller ovals lie inside the parallelogram oval because every rhombus and every rectangle is a parallelogram.

The two smaller ovals overlap in a lens-shaped region — this overlap represents quadrilaterals that are both a rectangle and a rhombus (i.e. all sides equal and all angles $$90^{\circ}$$), which is exactly a square.

A square is at the same time a rectangle (all angles $$90^{\circ}$$) and a rhombus (all sides equal). So the set of squares lies exactly in the region where the “Rectangles” oval and the “Rhombuses” oval overlap, inside the big “Parallelograms” oval.

No, not in general. On measuring the diagonals of the rhombus $$ABCD$$ constructed with $$\angle A = 60^{\circ}$$ and side $$4 \, \mathrm{cm}$$, we get $$AC = 4 \, \mathrm{cm}$$ (the shorter diagonal, spanning the two $$60^{\circ}$$ angles) and $$BD \approx 6.93 \, \mathrm{cm}$$ (the longer diagonal, spanning the two $$120^{\circ}$$ angles). These are unequal.

The diagonals of a rhombus are equal only in the special case when all four angles are $$90^{\circ}$$ — i.e. when the rhombus is a square.

Yes — the diagonals of a rhombus always intersect at $$90^{\circ}$$.

Reasoning. Let $$ABCD$$ be a rhombus with diagonals meeting at $$O$$. Since a rhombus is a parallelogram, its diagonals bisect each other, so $$OA = OC$$ and $$OB = OD$$.

Now compare $$\triangle AOB$$ and $$\triangle COB$$:

• $$AB = CB$$ (sides of the rhombus),
• $$OA = OC$$ (proved above),
• $$OB$$ is common.

By SSS, $$\triangle AOB \cong \triangle COB$$. Corresponding angles give $$\angle AOB = \angle COB$$. But these two angles form a linear pair along $$AC$$, so $$\angle AOB + \angle COB = 180^{\circ}$$. Hence each equals $$90^{\circ}$$.

So the diagonals of every rhombus meet at right angles.

Let $$O$$ be the point where the diagonals of the rhombus $$GAME$$ intersect. Since the diagonals of a rhombus (a parallelogram) bisect each other, $$GO = MO$$.

Now consider $$\triangle GEO$$ and $$\triangle MEO$$:

• $$GE = ME$$ (sides of the rhombus),
• $$GO = MO$$ (diagonals bisect each other, so the diagonal $$GM$$ is bisected at $$O$$),
• $$EO$$ is common.

By the SSS congruence rule, $$\triangle GEO \cong \triangle MEO$$.

The quadrilateral obtained is a square.

Justification. The two rubber-band diagonals have:

• equal length,
• they bisect each other (since they are placed to cross at their midpoints on the geoboard), and
• they are perpendicular to each other ($$90^{\circ}$$ between them).

Equal + bisect-each-other ⇒ rectangle. Add “perpendicular diagonals” ⇒ all sides are equal too. A rectangle with all sides equal is a square. Hence the quadrilateral obtained is a square.

Say we started with a square whose diagonals were of length $$d$$. When we extend one diagonal by $$2 \, \mathrm{cm}$$ on both sides, the two diagonals are still

• bisecting each other (the added lengths are equal on the two sides), and
• perpendicular (only the length changed, not the direction),

but they are no longer of equal length (one is now $$d + 4 \, \mathrm{cm}$$, the other is still $$d$$).

A quadrilateral whose diagonals bisect each other at right angles is a rhombus. It is not a square because the diagonals are now of different lengths. So the resulting quadrilateral is a rhombus (not a square).

Yes. Place the two equilateral triangles so that one side of the first triangle exactly matches one side of the second triangle, and glue them along that common edge. Since every side of every equilateral triangle has the same length ($$8 \, \mathrm{cm}$$), any pair of sides can be used as the common edge — the two triangles fit together perfectly. The resulting figure is bounded by the four remaining sides, so it is a quadrilateral.

Let the two equilateral triangles be $$\triangle ABD$$ and $$\triangle CBD$$, glued along the common side $$BD$$, forming the quadrilateral $$ABCD$$. The four sides are $$AB, BC, CD, DA$$, each equal to $$8 \, \mathrm{cm}$$ (since every side of an equilateral triangle of side $$8 \, \mathrm{cm}$$ is $$8 \, \mathrm{cm}$$).

So all four sides of $$ABCD$$ are equal ($$= 8 \, \mathrm{cm}$$). A quadrilateral with all sides equal is a rhombus.

The interior angles at $$B$$ and $$D$$ are each made of two $$60^{\circ}$$ angles of the equilateral triangles, so $$\angle B = \angle D = 120^{\circ}$$. The angles at $$A$$ and $$C$$ are $$60^{\circ}$$ each (one angle of an equilateral triangle). These are not $$90^{\circ}$$, so the rhombus is not a square.

To form a quadrilateral by joining two triangles along a matching side, the shared side must have the same length in both triangles. The two isosceles triangles each have sides $$8 \, \mathrm{cm}, 8 \, \mathrm{cm}$$ and $$6 \, \mathrm{cm}$$, so the possible common edges are:

1. Joining along an $$8 \, \mathrm{cm}$$ side of each triangle. Even within this case, the two triangles can be placed with the equal sides ($$8, 8$$) on the same side of the common edge (mirror image) or on opposite sides (rotated), giving two visually distinct quadrilaterals. Both have side lengths $$\{8, 6, 8, 6\}$$ or $$\{8, 8, 6, 6\}$$ depending on the orientation.
2. Joining along the $$6 \, \mathrm{cm}$$ side of each triangle. Only one arrangement — the two $$8 \, \mathrm{cm}$$ sides of each triangle become the remaining sides of the quadrilateral, giving all four remaining sides equal to $$8 \, \mathrm{cm}$$.

So there are essentially three different quadrilaterals (a kite, a parallelogram, and a rhombus) — see the next question for their identification.

Using the three arrangements identified in Question 50:

1. Join along an $$8 \, \mathrm{cm}$$ side, mirror image (both $$6 \, \mathrm{cm}$$ sides on one end, both remaining $$8 \, \mathrm{cm}$$ sides on the other): the resulting quadrilateral has two adjacent sides of $$6 \, \mathrm{cm}$$ and two adjacent sides of $$8 \, \mathrm{cm}$$. This is a kite (two pairs of adjacent equal sides).
2. Join along an $$8 \, \mathrm{cm}$$ side, rotated (one $$6 \, \mathrm{cm}$$ side and one $$8 \, \mathrm{cm}$$ side meeting at each of the two ‘other’ vertices): the sides are $$6, 8, 6, 8$$ around, and each pair of opposite sides is equal and parallel. This is a parallelogram.
3. Join along the $$6 \, \mathrm{cm}$$ side (the only $$6 \, \mathrm{cm}$$ edge of each triangle): all four remaining sides are $$8 \, \mathrm{cm}$$. So it is a quadrilateral with all sides equal — a rhombus.

Justification of the parallelogram case. Label the joined quadrilateral $$ABCD$$ with $$AB = 6, BC = 8, CD = 6, DA = 8$$. In $$\triangle ABD$$ and $$\triangle CDB$$ we have $$AB = CD = 6$$, $$AD = CB = 8$$ and common side $$BD$$; by SSS, they are congruent, giving $$\angle ABD = \angle CDB$$, so $$AB \parallel CD$$; similarly $$AD \parallel BC$$. Hence it is a parallelogram.

Justification of the rhombus case. All four sides equal $$\Rightarrow$$ rhombus (in fact a parallelogram with equal adjacent sides).

Two triangles fit together along a common side only if that side has the same length in both. Since the triangles are scalene, each side length ($$6, 9, 12 \, \mathrm{cm}$$) appears exactly once in each triangle. So the possible common edges are:

1. Join along the $$6 \, \mathrm{cm}$$ side.
2. Join along the $$9 \, \mathrm{cm}$$ side.
3. Join along the $$12 \, \mathrm{cm}$$ side.

For each choice of common edge, the second triangle can additionally be flipped (mirror image) — so it can be placed in two positions: with equal sides matched (mirror) or with unequal sides swapped (rotated). This gives up to $$3 \times 2 = 6$$ different quadrilaterals in all.

Look at each of the six arrangements from Question 52.

• Mirror joins (the two triangles reflect through the common edge): the two sides of one triangle meeting the common edge equal, respectively, the corresponding two sides of the second triangle meeting the common edge (i.e. adjacent pairs are equal). This is exactly the definition of a kite. Since the triangles are scalene, each of the three mirror joins gives a distinct kite.
• Rotated joins (the two triangles are related by a $$180^{\circ}$$ turn about the midpoint of the common edge): opposite sides of the resulting quadrilateral are equal (each side of one triangle becomes the opposite side of the other). By SSS ($$\triangle$$ divided by the diagonal), one can show that both pairs of opposite sides are parallel. So each rotated join gives a parallelogram. Since the triangles are scalene, none of these parallelograms is a rectangle, rhombus, or square. The three rotated joins give three distinct parallelograms.

Hence, joining two scalene triangles produces six different quadrilaterals — three kites and three parallelograms.

Steps of construction for a trapezium $$ABCD$$ with $$AB \parallel CD$$:

1. Draw the longer base $$AB$$ of some length, say $$6 \, \mathrm{cm}$$.
2. At $$A$$, using a protractor, draw a ray making an angle, say $$70^{\circ}$$, with $$AB$$.
3. At $$B$$, draw another ray making some other angle, say $$60^{\circ}$$, with $$BA$$.
4. Mark $$D$$ on the ray from $$A$$ with $$AD = 3 \, \mathrm{cm}$$.
5. Through $$D$$, draw a line parallel to $$AB$$ (with a set-square and ruler), meeting the ray from $$B$$ at $$C$$.
6. $$ABCD$$ is the required trapezium.

On measuring the two base angles at $$A$$ and $$B$$ (the ones between the parallel side $$AB$$ and the non-parallel sides), we get $$\angle A = 70^{\circ}$$ and $$\angle B = 60^{\circ}$$ (values will depend on how you choose to construct).

Yes. In a trapezium $$ABCD$$ with $$AB \parallel CD$$, the non-parallel side $$AD$$ acts as a transversal cutting the two parallel lines. So co-interior angles are supplementary:

$$\angle A + \angle D = 180^{\circ} \implies \angle D = 180^{\circ} - \angle A.$$

Similarly, $$BC$$ is a transversal, giving

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

Using the base angles from Question 55 ($$\angle A = 70^{\circ}, \angle B = 60^{\circ}$$),

$$\angle D = 180^{\circ} - 70^{\circ} = 110^{\circ},$$

$$\angle C = 180^{\circ} - 60^{\circ} = 120^{\circ}.$$

An isosceles trapezium is a trapezium in which the two non-parallel sides are equal in length. So the construction is the same as an ordinary trapezium, but with the two base angles at $$A$$ and $$B$$ chosen equal.

Steps:

1. Draw the longer base $$AB$$ of a chosen length.
2. At $$A$$ and at $$B$$, draw rays making equal angles with $$AB$$ (both angles pointing inward). Call the common angle $$\theta$$.
3. Mark $$D$$ on the ray from $$A$$ and $$C$$ on the ray from $$B$$ such that $$AD = BC$$.
4. Join $$DC$$. Then $$AB \parallel DC$$ automatically (because the two rays make equal angles with $$AB$$ and $$DC$$).

Alternatively: draw $$AB$$, draw a line parallel to $$AB$$ at the required height, and mark $$D$$ and $$C$$ on this parallel line symmetrically about the perpendicular bisector of $$AB$$. Joining them gives an isosceles trapezium.

Sample construction (take $$UV = 8 \, \mathrm{cm}$$ as the longer base, $$XW = 4 \, \mathrm{cm}$$, and equal non-parallel sides $$UX = VW = 3 \, \mathrm{cm}$$):

1. Draw $$UV = 8 \, \mathrm{cm}$$.
2. Locate the midpoint $$M$$ of $$UV$$.
3. Draw the perpendicular to $$UV$$ at $$M$$; on this perpendicular mark a segment $$XW$$ (of length $$4 \, \mathrm{cm}$$) symmetric about $$M$$ and at some fixed height above $$UV$$. That is, mark points $$X'$$ (above $$U$$'s side) and $$W'$$ (above $$V$$'s side) such that $$X'W' = 4 \, \mathrm{cm}$$ and $$M$$ is the midpoint of $$X'W'$$.
4. Adjust the height until $$UX' = VW' = 3 \, \mathrm{cm}$$ (use compass arcs of radius $$3 \, \mathrm{cm}$$ from $$U$$ and $$V$$, intersecting the horizontal line at $$X$$ and $$W$$).
5. Join $$UX, XW, WV$$. Then $$UVWX$$ is the required isosceles trapezium.

On measuring $$\angle U$$ with a protractor (using the sample values above), we get approximately $$\angle U \approx 48^{\circ}$$. In general the exact value depends on the chosen dimensions; for the sample, $$\cos \angle U = \dfrac{(UV - XW)/2}{UX} = \dfrac{2}{3}$$, so $$\angle U \approx 48^{\circ}$$.

In the isosceles trapezium $$UVWX$$ (with $$UV \parallel XW$$), drop perpendiculars from $$X$$ and $$W$$ to $$UV$$, meeting $$UV$$ at $$Y$$ and $$Z$$ respectively.

Then $$XY \perp UV$$ and $$WZ \perp UV$$, so $$XY \parallel WZ$$. Also, $$XW \parallel YZ$$ (both are horizontal — $$XW$$ by assumption and $$YZ$$ is a part of $$UV$$).

So $$XWZY$$ has both pairs of opposite sides parallel — it is a parallelogram. Moreover, its angles at $$Y$$ and $$Z$$ are $$90^{\circ}$$ (by construction), so all four angles are $$90^{\circ}$$. Hence $$XWZY$$ is a rectangle.

In $$\triangle UXY$$ and $$\triangle VWZ$$:

• $$UX = VW$$ (equal non-parallel sides of the isosceles trapezium),
• $$XY = WZ$$ (opposite sides of rectangle $$XWZY$$),
• $$\angle UYX = \angle VZW = 90^{\circ}$$ (both perpendiculars to $$UV$$).

By the RHS (right-angle–hypotenuse–side) congruence rule, $$\triangle UXY \cong \triangle VWZ$$.

A consequence: the corresponding parts give $$UY = VZ$$ and $$\angle YUX = \angle ZVW$$, i.e. the base angles $$\angle U$$ and $$\angle V$$ of the isosceles trapezium are equal.

(i) Rectangle $$ABCD$$ with $$\angle DBA = 30^{\circ}$$.

Let the diagonals meet at $$O$$. In a rectangle the diagonals are equal and bisect each other, so $$OA = OB = OC = OD$$.

All four angles of the rectangle are $$90^{\circ}$$. At vertex $$B$$: $$\angle ABC = 90^{\circ}$$ splits into $$\angle ABD + \angle DBC = 90^{\circ}$$. Given $$\angle ABD = 30^{\circ}$$, we get $$\angle DBC = 60^{\circ}$$.

In $$\triangle AOB$$, $$OA = OB$$, so it is isosceles with base angles $$\angle OAB = \angle OBA = 30^{\circ}$$. Then $$\angle AOB = 180^{\circ} - 30^{\circ} - 30^{\circ} = 120^{\circ}$$. Vertically opposite: $$\angle COD = 120^{\circ}$$. Linear pair: $$\angle BOC = \angle AOD = 60^{\circ}$$.

At vertex $$A$$: $$\angle DAB = 90^{\circ}$$ splits by diagonal $$AC$$ into $$\angle DAC + \angle CAB = 90^{\circ}$$. In isosceles $$\triangle AOB$$, $$\angle CAB = \angle OAB = 30^{\circ}$$, so $$\angle DAC = 60^{\circ}$$.

At vertex $$C$$: by symmetry, $$\angle BCA = 30^{\circ}$$ and $$\angle DCA = 60^{\circ}$$.

At vertex $$D$$: $$\angle ADB = 60^{\circ}$$ and $$\angle BDC = 30^{\circ}$$ (alternate interior angles / isosceles $$\triangle AOD$$).

(ii) Rectangle $$PQRS$$ with $$\angle QOR = 110^{\circ}$$.

Vertically opposite angles at $$O$$: $$\angle POS = 110^{\circ}$$. Linear pair: $$\angle POQ = \angle ROS = 180^{\circ} - 110^{\circ} = 70^{\circ}$$.

In $$\triangle POQ$$, $$OP = OQ$$ (diagonals of a rectangle bisect each other into equal halves), so it is isosceles. The base angles are $$\dfrac{180^{\circ} - 70^{\circ}}{2} = 55^{\circ}$$, i.e. $$\angle OPQ = \angle OQP = 55^{\circ}$$.

In $$\triangle QOR$$ (isosceles, $$OQ = OR$$), the base angles are $$\dfrac{180^{\circ} - 110^{\circ}}{2} = 35^{\circ}$$, i.e. $$\angle OQR = \angle ORQ = 35^{\circ}$$.

By symmetry, $$\angle ORS = \angle OSR = 55^{\circ}$$ and $$\angle OSP = \angle OPS = 35^{\circ}$$. As a check: $$\angle P = \angle OPQ + \angle OPS = 55^{\circ} + 35^{\circ} = 90^{\circ}$$ ✓.

Since $$PL$$ and $$AM$$ are diameters of the same circle, they have the same length:

$$PL = AM.$$

Both diameters pass through the centre $$O$$, and $$O$$ is the midpoint of each (a diameter is bisected by the centre). So the two diameters are two segments of equal length that bisect each other. In addition, they are perpendicular (given).

Now consider the quadrilateral $$APML$$ with vertices at the four endpoints (taken in the order around the circle: $$A, P, M, L$$). Its diagonals are $$AM$$ and $$PL$$ — precisely the two diameters — which are equal in length, bisect each other, and are perpendicular.

A quadrilateral whose diagonals are equal, bisect each other and are perpendicular is a square. Hence $$APML$$ is a square.

(You can also see this by symmetry: rotating the figure through $$90^{\circ}$$ about $$O$$ carries $$A \to P \to M \to L \to A$$, so all four sides are equal and all four angles are $$90^{\circ}$$.)

Use the diagonal property of a square. Steps:

1. Locate the midpoint of each stick (fold or measure).
2. Cross the two sticks at their midpoints — i.e. they meet at their own centres.
3. Adjust the sticks until the four endpoints all lie on a common circle (equivalently, until the thread, when wrapped around all four endpoints, forms a rhombus with all four sides equal).
4. Fix the sticks in this position.

Now the two sticks are diagonals of a quadrilateral that are equal (same length), bisect each other (crossed at midpoints) and give four equal sides (checked using the thread). By the reasoning of Question 3, this quadrilateral is a square, and its diagonals are perpendicular. So the angle between the sticks is exactly $$90^{\circ}$$.

Alternatively: cross the two sticks at their midpoints and use the thread to check that each of the two sticks divides the quadrilateral into two congruent triangles. When the four thread-sides are all equal, the sticks are perpendicular.

No. A quadrilateral in which both pairs of opposite sides are parallel (and hence equal) is a parallelogram, not necessarily a rectangle.

Counterexample. The parallelogram we constructed in Question 29 has $$AB = CD = 5 \, \mathrm{cm}$$, $$AD = BC = 4 \, \mathrm{cm}$$, $$AB \parallel CD$$ and $$AD \parallel BC$$, but its angles are $$30^{\circ}, 150^{\circ}, 30^{\circ}, 150^{\circ}$$ — clearly not $$90^{\circ}$$. So it is not a rectangle.

Hence “opposite sides parallel and equal” is not a definition of a rectangle. To get a rectangle we need one extra condition, e.g. one angle equal to $$90^{\circ}$$ (which forces all four to be $$90^{\circ}$$), or the diagonals equal.

Key property: the diagonals of a parallelogram bisect each other. So the two diagonals must cross at their midpoints — the halves are $$\dfrac{7}{2} = 3.5 \, \mathrm{cm}$$ and $$\dfrac{5}{2} = 2.5 \, \mathrm{cm}$$.

Steps of construction:

1. Draw a segment $$AC = 7 \, \mathrm{cm}$$ and mark its midpoint $$O$$ (so $$OA = OC = 3.5 \, \mathrm{cm}$$).
2. At $$O$$, using a protractor, draw a line making an angle of $$140^{\circ}$$ with $$AC$$.
3. On this line, mark $$B$$ and $$D$$ on opposite sides of $$O$$ such that $$OB = OD = 2.5 \, \mathrm{cm}$$ (so $$BD = 5 \, \mathrm{cm}$$).
4. Join $$AB, BC, CD, DA$$.

The quadrilateral $$ABCD$$ has diagonals $$AC = 7 \, \mathrm{cm}$$ and $$BD = 5 \, \mathrm{cm}$$ meeting at $$O$$ (their common midpoint) at $$140^{\circ}$$. Since the diagonals bisect each other, $$ABCD$$ is a parallelogram. Measuring, its sides are approximately $$AB = DC \approx 5.71 \, \mathrm{cm}$$ and $$BC = AD \approx 2.16 \, \mathrm{cm}$$.

Key property: the diagonals of a rhombus bisect each other at right angles. So the diagonals cross at their midpoints, and the angle between them is $$90^{\circ}$$.

Steps of construction:

1. Draw $$AC = 5 \, \mathrm{cm}$$ and mark its midpoint $$O$$ (so $$OA = OC = 2.5 \, \mathrm{cm}$$).
2. At $$O$$, draw a perpendicular to $$AC$$.
3. On this perpendicular, mark $$B$$ and $$D$$ on opposite sides of $$O$$ with $$OB = OD = 2 \, \mathrm{cm}$$ (so $$BD = 4 \, \mathrm{cm}$$).
4. Join $$AB, BC, CD, DA$$.

Since the diagonals bisect each other at $$90^{\circ}$$, the quadrilateral $$ABCD$$ is a rhombus. Its side is $$\sqrt{OA^2 + OB^2} = \sqrt{2.5^2 + 2^2} = \sqrt{10.25} \approx 3.20 \, \mathrm{cm}$$.

Take two equilateral triangles $$\triangle ABD$$ and $$\triangle CBD$$, each with side $$4 \, \mathrm{cm}$$, and glue them along the common side $$BD$$. The resulting quadrilateral is $$ABCD$$.

Sides. The four sides are $$AB, BC, CD, DA$$. Each of these is a side of one of the equilateral triangles, so

$$AB = BC = CD = DA = 4 \, \mathrm{cm}.$$

Angles. Each interior angle of an equilateral triangle is $$60^{\circ}$$. So:

At vertex $$A$$: only one triangle contributes, giving $$\angle A = 60^{\circ}$$.

At vertex $$C$$: similarly $$\angle C = 60^{\circ}$$.

At vertex $$B$$: both triangles contribute, so $$\angle B = 60^{\circ} + 60^{\circ} = 120^{\circ}$$.

At vertex $$D$$: similarly $$\angle D = 120^{\circ}$$.

Check: $$60^{\circ}+120^{\circ}+60^{\circ}+120^{\circ} = 360^{\circ}$$ ✓.

The quadrilateral has all sides equal and unequal adjacent angles — it is a rhombus (not a square).

Recall (from Question 54) that in a kite $$ABCD$$ (with $$AB = BC$$ and $$CD = DA$$), the diagonal $$BD$$ is the axis of symmetry: it is perpendicular to the other diagonal $$AC$$ and bisects it. So $$AC$$ is bisected by $$BD$$, but $$BD$$ itself need not be bisected by $$AC$$ (they meet at some interior point $$O$$ with $$AO = OC$$, but $$BO$$ and $$OD$$ can be different).

Steps of construction (take $$BD = 8 \, \mathrm{cm}$$ as the axis of symmetry and $$AC = 6 \, \mathrm{cm}$$ as the other diagonal, with $$O$$ splitting $$BD$$ into $$OB = 3 \, \mathrm{cm}$$ and $$OD = 5 \, \mathrm{cm}$$, say):

1. Draw segment $$BD = 8 \, \mathrm{cm}$$.
2. On $$BD$$, choose an interior point $$O$$ (say at $$3 \, \mathrm{cm}$$ from $$B$$; you may pick a different split).
3. At $$O$$, draw a line perpendicular to $$BD$$.
4. On the perpendicular, mark $$A$$ and $$C$$ on opposite sides of $$O$$ with $$OA = OC = 3 \, \mathrm{cm}$$ (so $$AC = 6 \, \mathrm{cm}$$).
5. Join $$AB, BC, CD, DA$$.

Then $$ABCD$$ is a kite. By construction, $$OA = OC$$, so $$\triangle AOB \cong \triangle COB$$ (SAS with common $$OB$$ and right angles at $$O$$), giving $$AB = CB$$; similarly $$AD = CD$$. So $$ABCD$$ has two pairs of adjacent equal sides — a kite.

From the figure, $$R$$ is the common vertex of the two rectangles. In rectangle $$PAIR$$, the diagonal $$PI$$ passes through $$R$$-region, and the angle marked $$30^{\circ}$$ at $$R$$ is the angle $$\angle IRP$$ that the diagonal from $$R$$ makes with side $$RP$$. Since $$PR = 5 \, \mathrm{cm}$$ and the two rectangles are congruent (also $$RS = 5 \, \mathrm{cm}$$), the corresponding diagonal in $$RODS$$ makes the same $$30^{\circ}$$ angle with its side.

In rectangle $$PAIR$$ let the diagonals $$PI$$ and $$AR$$ meet at their common midpoint (call it $$M$$). Since the diagonals of a rectangle are equal and bisect each other, $$\triangle MRP$$ is isosceles with $$MR = MP$$, so the base angles are equal: $$\angle MPR = \angle MRP = 30^{\circ}$$. Hence $$\angle PMR = 180^{\circ} - 30^{\circ} - 30^{\circ} = 120^{\circ}$$, so the diagonals of $$PAIR$$ meet at $$120^{\circ}$$ and $$60^{\circ}$$.

Now, apply the same reasoning to rectangle $$RODS$$. Its diagonals $$RD$$ and $$OS$$ meet at their common midpoint. By the same $$30^{\circ}$$ argument on the corresponding side, the diagonals of $$RODS$$ also meet at $$60^{\circ}$$ and $$120^{\circ}$$.

Since $$O$$ is a vertex of rectangle $$RODS$$ and $$I$$ lies collinearly with $$R$$ and $$D$$ on the shared bottom line, the angle $$\angle IOD$$ is exactly the base angle of the isosceles triangle formed at $$O$$ by half-diagonals — and by the same $$30^{\circ}$$ analysis,

$$\angle IOD = 30^{\circ}.$$

A square is characterised by its two diagonals being (i) equal, (ii) bisecting each other, and (iii) perpendicular. We can construct the perpendicular bisector of a segment using only a compass and a ruler.

Steps of construction:

1. Draw the segment $$AC = 6 \, \mathrm{cm}$$.
2. With centre $$A$$ and any radius greater than $$3 \, \mathrm{cm}$$, draw arcs above and below $$AC$$.
3. With centre $$C$$ and the same radius, draw two more arcs; they cut the first arcs at two points. The line joining these two points is the perpendicular bisector of $$AC$$; it meets $$AC$$ at the midpoint $$O$$ (with $$OA = OC = 3 \, \mathrm{cm}$$) and is perpendicular to $$AC$$.
4. On this perpendicular bisector, mark two points $$B$$ and $$D$$ on opposite sides of $$O$$, each at distance $$3 \, \mathrm{cm}$$ from $$O$$ (use the compass with radius $$3 \, \mathrm{cm}$$ centred at $$O$$).
5. Join $$AB, BC, CD, DA$$.

The two diagonals $$AC$$ and $$BD$$ are equal ($$6 \, \mathrm{cm}$$ each), bisect each other at $$O$$, and are perpendicular (by construction). Hence $$ABCD$$ is a square. On measuring, each side is $$3\sqrt{2} \approx 4.24 \, \mathrm{cm}$$.

Let the square $$CASE$$ have side $$s$$, and let $$U, V, W, X$$ be the midpoints of sides $$CA, AS, SE, EC$$ respectively.

Sides of $$UVWX$$. Consider $$\triangle AUV$$: it is a right triangle at $$A$$ with legs $$AU = AV = s/2$$. So

$$UV = \sqrt{(s/2)^2 + (s/2)^2} = \dfrac{s}{\sqrt{2}}.$$

By symmetry (the same argument at each corner of $$CASE$$), all four sides of $$UVWX$$ equal $$s/\sqrt{2}$$. So $$UVWX$$ has all four sides equal.

Angles of $$UVWX$$. $$\triangle AUV$$ is an isosceles right triangle at $$A$$, so $$\angle AVU = 45^{\circ}$$; similarly $$\angle SVW = 45^{\circ}$$. The straight angle at $$V$$ along $$AS$$ is $$180^{\circ}$$, so

$$\angle UVW = 180^{\circ} - 45^{\circ} - 45^{\circ} = 90^{\circ}.$$

By symmetry, all four angles of $$UVWX$$ are $$90^{\circ}$$.

Since all sides are equal and all angles are $$90^{\circ}$$, $$UVWX$$ is a square.

Construction and measurement. Draw square $$CASE$$ (say side $$4 \, \mathrm{cm}$$). Mark midpoints of its sides using a ruler. Join them. On measurement, all sides come out equal ($$\approx 2.83 \, \mathrm{cm} = 4/\sqrt{2}$$) and all angles $$90^{\circ}$$.

Other inner squares (Fig (b)). Instead of taking midpoints, on each side of $$CASE$$ mark a point at the same distance $$k$$ from one endpoint (say $$k$$ from $$C$$ on $$CA$$, $$k$$ from $$A$$ on $$AS$$, $$k$$ from $$S$$ on $$SE$$, $$k$$ from $$E$$ on $$EC$$). By a similar SAS argument (each corner triangle is a right triangle with legs $$k$$ and $$s-k$$), the inner quadrilateral has all four sides equal and all four angles $$90^{\circ}$$ — it is again a square. Varying $$k$$ from $$0$$ to $$s$$ gives infinitely many such inner squares (the midpoint case $$k = s/2$$ produces the smallest inner square).

Yes. Let $$ABCD$$ have $$AB = BC = CD = DA = s$$ and $$\angle A = 90^{\circ}$$.

Since all four sides are equal, $$ABCD$$ is a rhombus, and hence a parallelogram. In a parallelogram, opposite angles are equal and adjacent angles are supplementary. So

$$\angle C = \angle A = 90^{\circ}, \qquad \angle B = 180^{\circ} - \angle A = 90^{\circ}, \qquad \angle D = 90^{\circ}.$$

Hence all four angles are $$90^{\circ}$$ and all four sides are equal — the definition of a square.

By construction. Draw $$AB = 4 \, \mathrm{cm}$$. At $$A$$, construct a $$90^{\circ}$$ angle and mark $$D$$ with $$AD = 4 \, \mathrm{cm}$$. With centre $$B$$ and radius $$4 \, \mathrm{cm}$$, draw an arc; with centre $$D$$ and radius $$4 \, \mathrm{cm}$$, draw another arc; they meet at $$C$$. Measuring shows $$BC = DC = 4 \, \mathrm{cm}$$ and all angles of $$ABCD$$ are $$90^{\circ}$$ — a square.

Such a quadrilateral is a parallelogram. That is, if $$ABCD$$ has $$AB = CD$$ and $$AD = BC$$, then $$AB \parallel CD$$ and $$AD \parallel BC$$.

Proof. Draw the diagonal $$BD$$. In $$\triangle ABD$$ and $$\triangle CDB$$:

• $$AB = CD$$ (given),
• $$AD = CB$$ (given),
• $$BD = DB$$ (common side).

By SSS, $$\triangle ABD \cong \triangle CDB$$.

Corresponding angles give $$\angle ABD = \angle CDB$$. These are alternate interior angles at the transversal $$BD$$ cutting lines $$AB$$ and $$CD$$. Alternate interior angles being equal ⇒ the lines are parallel:

$$AB \parallel CD.$$

Similarly, $$\angle ADB = \angle CBD$$ (alternate angles at $$BD$$ cutting $$AD$$ and $$BC$$) ⇒ $$AD \parallel BC$$.

So both pairs of opposite sides are parallel, i.e. $$ABCD$$ is a parallelogram.

Yes. The angle sum of every quadrilateral — convex or concave — is $$360^{\circ}$$, provided we measure the interior angles (using the reflex angle at a concave vertex).

Reason. Draw a diagonal that lies entirely inside the quadrilateral (in a concave quadrilateral, one of the two possible diagonals does lie inside; the other stays outside). This diagonal splits the quadrilateral into two triangles.

Each triangle has angle sum $$180^{\circ}$$. Adding these two sums gives $$360^{\circ}$$. When we do the addition, the two ‘pieces’ of the interior angle at the concave vertex re-combine to give the full reflex interior angle. So the sum of the four interior angles of the quadrilateral is

$$180^{\circ} + 180^{\circ} = 360^{\circ}.$$

By construction. On drawing such a concave quadrilateral, on measuring the three non-reflex angles and the reflex angle at $$D$$ with a protractor (careful to measure the interior, i.e. reflex, angle at $$D$$), the total comes out to $$360^{\circ}$$.