Chapter 6: Triangles

Given figures

• PQRS is a rhombus; each side is $$1.5\,\text{cm}$$.
• ABCD is a square; each side is $$3\,\text{cm}$$.

Condition for similarity of two quadrilaterals
Two quadrilaterals are similar when both of the following hold:

1. Their corresponding angles are equal.
2. The ratios of their corresponding sides are equal (i.e. proportional).

Checking the side–ratio condition

Every side of PQRS measures $$1.5\,\text{cm}$$ and every side of ABCD measures $$3\,\text{cm}$$, so for any pair of corresponding sides we have
$$\dfrac{\text{side of PQRS}}{\text{side of ABCD}}=\dfrac{1.5}{3}=\dfrac12$$.
Thus all corresponding sides are indeed proportional.

Checking the angle condition

• In a square, each interior angle equals $$90^\circ$$.
• In a rhombus, the opposite angles are equal, but adjacent angles are supplementary: if one angle is $$\theta$$, the next is $$180^\circ-\theta$$. Only when $$\theta=90^\circ$$ does the rhombus become a square.

The diagram (Fig. 6.8) shows PQRS as an ordinary rhombus (its angles are not right angles). Hence at least one pair of corresponding angles in PQRS and ABCD are unequal.

Conclusion
Although their sides are proportional, their corresponding angles are not equal, so the quadrilaterals fail the angle condition. Therefore PQRS and ABCD are not similar.

Given: In $$\triangle ABC$$, a line through D on AB and E on AC is drawn such that $$DE \parallel BC$$.

To prove: $$\dfrac{AD}{AB}=\dfrac{AE}{AC}$$.

Construction: (Already done in the statement.)
For clarity, notice that points D and E lie on AB and AC respectively, and segment DE is parallel to BC.

Proof:

1. Because $$DE \parallel BC$$, the following pairs of angles are equal (corresponding angles):
      $$\angle ADE = \angle ABC$$  and  $$\angle AED = \angle ACB$$.
2. Also, $$\angle DAE$$ is common to both $$\triangle ADE$$ and $$\triangle ABC$$.
3. Thus the two triangles are similar by the AAA similarity criterion:
      $$\triangle ADE \sim \triangle ABC$$.
4. For similar triangles, the ratios of corresponding sides are equal. The correspondence is    $$AD \leftrightarrow AB$$, $$AE \leftrightarrow AC$$ and $$DE \leftrightarrow BC$$.
   Therefore,
      $$\dfrac{AD}{AB}=\dfrac{AE}{AC}=\dfrac{DE}{BC}$$.
5. Taking the first two equal fractions gives the required relation:
      $$\dfrac{AD}{AB}=\dfrac{AE}{AC}$$.

Hence proved.

Step 1 : Identify the three parallel lines

We are told that $$\mathrm{AB} \parallel \mathrm{DC}$$ and that $$\mathrm{EF} \parallel \mathrm{AB}$$.
Hence all three lines are parallel:

$$\mathrm{AB} \;\parallel\; \mathrm{EF} \;\parallel\; \mathrm{DC}$$

Step 2 : Locate the two transversals

The non-parallel sides $$\mathrm{AD}$$ and $$\mathrm{BC}$$ cut the three parallel lines.
Thus

• on transversal $$\mathrm{AD}$$ the points of intersection are $$A,E,D$$, and
• on transversal $$\mathrm{BC}$$ the points of intersection are $$B,F,C$$.

Step 3 : Apply the Intercept (Basic Proportionality) Theorem

The theorem states: “If three (or more) parallel lines cut two transversals, then they cut the transversals in proportional segments.” Therefore

$$\dfrac{\mathrm{AE}}{\mathrm{ED}} = \dfrac{\mathrm{BF}}{\mathrm{FC}}$$

Step 4 : Conclude

Hence the required equality is established:

$$\boxed{\dfrac{\mathrm{AE}}{\mathrm{ED}} = \dfrac{\mathrm{BF}}{\mathrm{FC}}}$$

Given : In ΔPQR, point S lies on PQ and point T lies on PR such that

$$\dfrac{PS}{SQ}=\dfrac{PT}{TR} \quad \text{and} \quad \angle PST = \angle PRQ$$

To prove : ΔPQR is isosceles (i.e. $$PQ = PR$$).

1. Proving ST ∥ QR

In ΔPQR the points S and T divide the sides PQ and PR in the same ratio:

$$\dfrac{PS}{SQ}=\dfrac{PT}{TR}.$$

By the Converse of the Basic Proportionality Theorem (if a line through two sides of a triangle divides them in the same ratio, that line is parallel to the third side), we get

$$ST \;\parallel\; QR\;.$$

2. Converting the given angle equality by using the parallelism

Because $$ST \parallel QR$$, the angle between PS and ST equals the angle between the lines PS (or PQ, because S lies on PQ) and QR. Therefore

$$\angle PST = \angle(PS,ST)=\angle(PS,QR)=\angle PQR.$$

But it is given that $$\angle PST = \angle PRQ$$. Hence

$$\angle PQR = \angle PRQ.$$

3. Concluding that ΔPQR is isosceles

In a triangle, equal angles lie opposite equal sides. The equal angles just obtained are opposite the sides PR and PQ respectively. Therefore

$$PR = PQ.$$

Thus ΔPQR has two equal sides and is an isosceles triangle.

Hence proved.

Given data from Fig. 6.17 (i)
Triangle $$\triangle ABC$$ with the point $$D$$ on $$AB$$ and the point $$E$$ on $$AC$$ such that $$DE \parallel BC$$.
The lengths marked in the figure are

• $$AD = 1.5\text{ cm}$$
• $$DB = 3\text{ cm}$$  (so $$AB = AD + DB = 4.5\text{ cm}$$)
• $$AE = 1.5\text{ cm}$$

Step 1 Apply the Basic Proportionality Theorem (BPT)
Because $$DE \parallel BC$$, BPT gives $$\frac{AD}{DB}=\frac{AE}{EC}.$$

Step 2 Insert the known numbers
$$\frac{1.5}{3}=\frac{1.5}{EC}.$$

Step 3 Solve for $$EC$$
Cross-multiplying, $$1.5\,EC = 1.5\times 3 \;\Rightarrow\; EC = 3\text{ cm}.$$

Given data from Fig. 6.17 (ii)
Again $$DE \parallel BC$$ with the points $$D\in AB$$ and $$E\in AC$$.
The lengths marked are

• $$AB = 4.5\text{ cm}$$
• $$AE = 1.5\text{ cm}$$
• $$EC = 3\text{ cm}$$

Step 1 Apply BPT
$$\frac{AD}{DB}=\frac{AE}{EC}$$  ⇒  $$\dfrac{x}{4.5-x}=\dfrac{1.5}{3}=\dfrac12.$$

Step 2 Solve for $$x$$
Cross-multiplying, $$2x = 4.5 - x \;\Rightarrow\; 3x = 4.5 \;\Rightarrow\; x = 1.5.$$

Therefore $$AD = 1.5\text{ cm}.$$

Results
Fig. 6.17 (i): $$EC = 3\text{ cm}$$
Fig. 6.17 (ii): $$AD = 1.5\text{ cm}$$

Given. In the figure, $$\mathrm{LM} \parallel \mathrm{CB}$$ and $$\mathrm{LN} \parallel \mathrm{CD}$$. Here $$\mathrm{M}$$ lies on $$\mathrm{AB}$$, $$\mathrm{N}$$ lies on $$\mathrm{AD}$$ and $$\mathrm{L}$$ lies on $$\mathrm{AC}$$.

1. Similarity in $$\triangle \mathrm{ABC}$$.

• Since $$\mathrm{LM} \parallel \mathrm{CB}$$, corresponding angles are equal:
    $$\angle \mathrm{ALM} = \angle \mathrm{ACB}$$ and $$\angle \mathrm{AML} = \angle \mathrm{ABC}$$.
• Thus $$\triangle \mathrm{ALM} \sim \triangle \mathrm{ACB}$$ (AAA criterion).
• Corresponding sides of similar triangles are proportional, so
    $$\dfrac{\mathrm{AM}}{\mathrm{AB}} = \dfrac{\mathrm{AL}}{\mathrm{AC}} \qquad (1)$$

2. Similarity in $$\triangle \mathrm{ACD}$$.

• Because $$\mathrm{LN} \parallel \mathrm{CD}$$, we have
    $$\angle \mathrm{ALN} = \angle \mathrm{ACD}$$ and $$\angle \mathrm{ANL} = \angle \mathrm{ADC}$$.
• Therefore $$\triangle \mathrm{ALN} \sim \triangle \mathrm{ACD}$$ (AAA criterion).
• Hence
    $$\dfrac{\mathrm{AN}}{\mathrm{AD}} = \dfrac{\mathrm{AL}}{\mathrm{AC}} \qquad (2)$$

3. Comparing the two ratios.

From (1) and (2): $$\dfrac{\mathrm{AM}}{\mathrm{AB}} = \dfrac{\mathrm{AL}}{\mathrm{AC}} = \dfrac{\mathrm{AN}}{\mathrm{AD}}$$.

Hence $$\dfrac{\mathrm{AM}}{\mathrm{AB}} = \dfrac{\mathrm{AN}}{\mathrm{AD}}$$, as required.

Proved.

To draw the figure
Draw $$\triangle ABC$$ with base $$BC$$. Mark a point $$D$$ on side $$AB$$. Through $$D$$ draw a line parallel to $$AC$$; let it cut $$BC$$ at $$E$$. Now through the same point $$D$$ draw another line parallel to $$AE$$; let it meet $$BE$$ at $$F$$. The required configuration (Fig. 6.19) is obtained.

Goal  Show that $$\dfrac{BF}{FE}=\dfrac{BE}{EC}$$.

Step 1 : Using $$DE\,\parallel\,AC$$ in $$\triangle ABC$$

Since a line parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem, BPT), we have

$$\dfrac{BD}{DA}=\dfrac{BE}{EC}\;\;\;(1)$$

Step 2 : Using $$DF\,\parallel\,AE$$ in $$\triangle BAE$$

Again applying BPT, now to $$\triangle BAE$$ with $$DF\parallel AE$$, we get

$$\dfrac{BD}{DA}=\dfrac{BF}{FE}\;\;\;(2)$$

Step 3 : Comparing the two ratios

From (1) and (2), the common left-hand side $$\dfrac{BD}{DA}$$ gives

$$\dfrac{BF}{FE}=\dfrac{BE}{EC}$$

Hence the required relation is proved.

Given : In ΔPQR, point O is any point on side QR.
A point D is taken on side PQ.
Through D two lines are drawn so that DE ∥ OQ and DF ∥ OR. The two parallels cut PR in E and QR (or its extension) in F, respectively. We have to prove that EF ∥ QR.

Proof :

1. In ΔPOQ, the line DE is drawn parallel to OQ and meets the other two sides of the triangle at D (on PQ) and E (on PO). By the Basic Proportionality Theorem (BPT), we get

$$\frac{PD}{DQ}=\frac{PE}{EO}\;\;\;\;\;(1)$$

2. In ΔPOR, the line DF is drawn parallel to OR and meets the other two sides of that triangle at D (on PQ) and F (on PR). Again using BPT,

$$\frac{PD}{DQ}=\frac{PF}{FR}\;\;\;\;(2)$$

3. From (1) and (2), the left–hand sides are equal, so the right–hand sides are equal as well:

$$\frac{PE}{EO}=\frac{PF}{FR}\;\;\;\;(3)$$

4. Combine each numerator and denominator in (3) to refer only to the sides of ΔPQR :

$$\frac{PE}{EQ}=\frac{PF}{FR}\;\;\;\;(4)$$

5. Relation (4) tells us that the points E and F divide the two sides PQ and PR of ΔPQR in the same ratio. By the converse of the Basic Proportionality Theorem, if a line joining two points on two sides of a triangle cuts those sides in the same ratio, that line is parallel to the third side. Hence

EF ∥ QR .

Thus, using the Basic Proportionality Theorem twice and its converse once, we have proved that EF is parallel to QR when DE ∥ OQ and DF ∥ OR.

Given data

• O is a point from which three rays $$\overline{OP}$$, $$\overline{OQ}$$ and $$\overline{OR}$$ are drawn so that $$P,Q,R$$ are non-collinear (hence $$\triangle PQR$$ is formed).
• A lies on $$\overline{OP}$$, B lies on $$\overline{OQ}$$ and C lies on $$\overline{OR}$$.
• $$\overline{AB} \parallel \overline{PQ}$$ and $$\overline{AC} \parallel \overline{PR}$$.

Required to prove : $$\overline{BC} \parallel \overline{QR}$$.

Step 1 : Use similarity produced by $$\overline{AB} \parallel \overline{PQ}$$.

In $$\triangle OPQ$$ the line $$\overline{AB}$$ is drawn through A on $$\overline{OP}$$ parallel to $$\overline{PQ}$$:

• $$\angle OAB = \angle OPQ$$ (alternate interior angles)
• $$\angle OBA = \angle OQP$$ (corresponding angles)

Therefore $$\triangle OAB \sim \triangle OPQ$$ (AA similarity).

Corresponding sides of similar triangles are proportional, so

$$\frac{OA}{OP} = \frac{OB}{OQ} = \frac{AB}{PQ} \;(1)$$

Step 2 : Use similarity produced by $$\overline{AC} \parallel \overline{PR}$$.

In $$\triangle OPR$$ the line $$\overline{AC}$$ is drawn through A on $$\overline{OP}$$ parallel to $$\overline{PR}$$:

• $$\angle OAC = \angle OPR$$ (alternate interior angles)
• $$\angle OCA = \angle ORP$$ (corresponding angles)

Hence $$\triangle OAC \sim \triangle OPR$$, and

$$\frac{OA}{OP} = \frac{OC}{OR} = \frac{AC}{PR} \;(2)$$

Step 3 : Obtain a proportion between the segments on $$\overline{OQ}$$ and $$\overline{OR}$$.

From (1) and (2) the first ratios are identical, so

$$\frac{OB}{OQ} = \frac{OA}{OP} = \frac{OC}{OR} \;\Rightarrow\; \frac{OB}{OQ} = \frac{OC}{OR} \;(3)$$

Step 4 : Apply the converse of the Basic Proportionality Theorem in $$\triangle OQR$$.

In $$\triangle OQR$$, points B and C lie on the sides $$\overline{OQ}$$ and $$\overline{OR}$$ respectively. Equation (3) shows that

$$\frac{OB}{OQ} = \frac{OC}{OR}.$$

The converse of the Basic Proportionality Theorem (Thales’ theorem) states:

• If a line through two points on the sides of a triangle divides those two sides in the same ratio, then that line is parallel to the third side.

Because the equal ratio condition is met, the line $$\overline{BC}$$ must be parallel to the third side $$\overline{QR}$$ of $$\triangle OQR$$.

Thus $$\overline{BC} \parallel \overline{QR}$$, as required.

Conclusion

The two pairs of given parallels lead to two pairs of similar triangles, from which equal ratios on sides $$\overline{OQ}$$ and $$\overline{OR}$$ are deduced. By the converse of the Basic Proportionality Theorem these equal ratios guarantee that $$\overline{BC}$$ is parallel to $$\overline{QR}$$.

Hence proved.

Given $$\triangle ABC$$. Let $$D$$ be the mid-point of side $$AB$$, so $$AD = DB$$.

Construction Through point $$D$$ draw a line $$DE$$ parallel to $$BC$$ which meets $$AC$$ at $$E$$.

To prove $$E$$ is the mid-point of $$AC$$, i.e. $$AE = EC$$.

Proof

1. Since $$DE \parallel BC$$ and it intersects the other two sides $$AB$$ and $$AC$$ at $$D$$ and $$E$$ respectively, we can apply Theorem 6.1 (Basic Proportionality Theorem).
2. By Theorem 6.1: $$\frac{AD}{DB} = \frac{AE}{EC}$$.
3. Because $$D$$ is the mid-point of $$AB$$, we have $$AD = DB$$, so $$\frac{AD}{DB} = 1$$.
4. Substituting in the equality from step 2 gives $$\frac{AE}{EC} = 1$$.
5. Hence $$AE = EC$$, showing that $$E$$ bisects $$AC$$.

Therefore the line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.  Proved.

Given : A triangle $$\triangle ABC$$ in which points $$D$$ and $$E$$ are respectively the mid-points of sides $$AB$$ and $$AC$$.

To prove : The segment $$DE$$ is parallel to the third side $$BC$$, i.e. $$DE \parallel BC$$.

Reasoning tool : Theorem 6.2 (Converse of the Basic Proportionality Theorem) — If a line drawn through a triangle cuts two of its sides proportionally, then the line is parallel to the third side.

Construction (only for clarity) : Join $$D$$ to $$E$$.

Proof :

1. Because $$D$$ is the mid-point of $$AB$$, we have $$AD = DB \implies \dfrac{AD}{DB} = 1.$$
2. Because $$E$$ is the mid-point of $$AC$$, we have $$AE = EC \implies \dfrac{AE}{EC} = 1.$$
3. From steps 1 and 2, the two ratios are equal: $$\dfrac{AD}{DB} = \dfrac{AE}{EC}.$$
4. The line $$DE$$ cuts sides $$AB$$ and $$AC$$ of $$\triangle ABC$$ at points $$D$$ and $$E$$ such that the division of the two sides is in the same ratio (step 3).
5. Therefore, by Theorem 6.2, the line $$DE$$ is parallel to the remaining side $$BC$$: $$DE \parallel BC.$$

Hence, the segment joining the mid-points of any two sides of a triangle is parallel to the third side. (Mid-Point Theorem)

Given: ABCD is a trapezium with $$\mathrm{AB} \parallel \mathrm{DC}$$. Diagonals $$\mathrm{AC}$$ and $$\mathrm{BD}$$ intersect at $$\mathrm{O}$$.

To prove: $$\dfrac{\mathrm{AO}}{\mathrm{BO}} = \dfrac{\mathrm{CO}}{\mathrm{DO}}$$.

Construction (for reference while reading): Draw the trapezium. Mark diagonals $$\mathrm{AC}$$ and $$\mathrm{BD}$$ so that they meet at $$\mathrm{O}$$.

Proof:

1. In $$\triangle \mathrm{AOB}$$ and $$\triangle \mathrm{COD}$$, note the following equal angles.
   
   • $$\angle \mathrm{AOB} = \angle \mathrm{COD}$$   (vertically opposite angles – the two diagonals cross at $$\mathrm{O}$$).
   • $$\angle \mathrm{ABO} = \angle \mathrm{CDO}$$   (alternate interior angles, as $$\mathrm{AB} \parallel \mathrm{DC}$$ and $$\mathrm{BD}$$ is a transversal).
   • $$\angle \mathrm{BAO} = \angle \mathrm{DCO}$$   (alternate interior angles, as $$\mathrm{AB} \parallel \mathrm{DC}$$ and $$\mathrm{AC}$$ is a transversal).
2. The two triangles therefore have two equal angles each; hence
   $$\triangle \mathrm{AOB} \sim \triangle \mathrm{COD}$$   (AAA similarity criterion).
3. Corresponding sides of similar triangles are proportional, so
   $$\dfrac{\mathrm{AO}}{\mathrm{CO}} \,=\, \dfrac{\mathrm{BO}}{\mathrm{DO}} \,=\, \dfrac{\mathrm{AB}}{\mathrm{DC}}.$$
4. Taking the first two ratios from step 3 and writing them in the desired order gives
   $$\dfrac{\mathrm{AO}}{\mathrm{BO}} = \dfrac{\mathrm{CO}}{\mathrm{DO}}.$$

The required relation is proved.

Given. In quadrilateral ABCD the diagonals AC and BD intersect at O such that

$$\dfrac{AO}{BO}=\dfrac{CO}{DO}$$

To prove. ABCD is a trapezium  (i.e. one pair of opposite sides is parallel).

Step 1 – Choose two triangles.
Consider triangles $$\triangle AOB$$ and $$\triangle COD$$.

(a) Equality of a pair of angles
Because the diagonals intersect, the vertical angle theorem gives

$$\angle AOB = \angle COD.$$

(b) Proportional pair of sides
From the hypothesis we already have

$$\dfrac{AO}{BO}=\dfrac{CO}{DO}.$$

Step 2 – Apply SAS similarity.
In $$\triangle AOB$$ and $$\triangle COD$$ we now know

• one pair of corresponding sides proportional: $$AO / BO = CO / DO$$,
• the included angles equal: $$\angle AOB = \angle COD$$.

Hence the two triangles are similar by the Side–Angle–Side (SAS) similarity criterion:

$$\triangle AOB \sim \triangle COD.$$

Step 3 – Deduce a pair of equal angles.
Similarity gives equality of corresponding angles. Matching the order of sides used above, vertex B of the first triangle corresponds to vertex D of the second, so

$$\angle ABO = \angle CDO.$$

Step 4 – Translate the angle equality into parallelism.
Line BD meets AB at B and CD at D, so BD is a transversal for the pair of lines AB and CD.
The equality $$\angle ABO = \angle CDO$$ represents a pair of corresponding angles with respect to this transversal. When corresponding angles are equal, the two lines are parallel. Therefore

$$AB \parallel CD.$$

Step 5 – Conclude.
A quadrilateral with one pair of opposite sides parallel is a trapezium. Since we have proved $$AB \parallel CD$$, the quadrilateral ABCD is a trapezium.

Hence proved.

Suggestion for a diagram. Draw quadrilateral ABCD with the longer diagonal AC and the shorter diagonal BD crossing at O. Mark the equal angle pairs and indicate the proportional segments on the two diagonals.

Given: In the figure, $$\mathrm{PQ} \parallel \mathrm{RS}$$ and the two straight lines $$\mathrm{PR}$$ and $$\mathrm{QS}$$ intersect each other at $$\mathrm{O}$$.

We have to prove that $$\triangle \mathrm{POQ} \sim \triangle \mathrm{SOR}$$.

Proof

1. First pair of equal angles
   Because $$\mathrm{PR}$$ and $$\mathrm{QS}$$ cross at $$\mathrm{O}$$, the vertically opposite angles are equal:
   $$\angle \mathrm{POQ}=\angle \mathrm{SOR}$$.

2. Second pair of equal angles
   Since $$\mathrm{PQ} \parallel \mathrm{RS}$$ and $$\mathrm{QS}$$ acts as a transversal, the following are corresponding angles and therefore equal:
   $$\angle \mathrm{PQO}=\angle \mathrm{OSR}$$.

3. Similarity by AA criterion
   In $$\triangle \mathrm{POQ}$$ and $$\triangle \mathrm{SOR}$$ we have
   • $$\angle \mathrm{POQ}=\angle \mathrm{SOR}$$ (step 1)
   • $$\angle \mathrm{PQO}=\angle \mathrm{OSR}$$ (step 2)
   Hence, by the AA (Angle–Angle) similarity criterion,
   $$\triangle \mathrm{POQ}\sim\triangle \mathrm{SOR}$$.

Thus the required similarity is established.

Unable to provide a worked solution because the figure (Fig. 6.30) referred to in the question is not available. Without the figure the data are insufficient to determine $$\angle P$$ unambiguously.

Given data
Two straight lines $$AB$$ and $$CD$$ intersect at $$O$$ in such a way that $$OA\,\cdot\,OB = OC\,\cdot\,OD$$.

To prove
$$\angle A = \angle C \;\text{and}\; \angle B = \angle D$$.

Construction
Through the non-collinear points $$A,\;B$$ and $$C$$ draw a circle. Let this circle meet the straight line $$CD$$ again at a point $$D'$$ (so $$D' \neq C$$).

Proof

1. Because $$A,\,B,\,C,\,D'$$ lie on the same circle, the chords $$AB$$ and $$CD'$$ intersect at $$O$$. By the Intersecting-Chords Theorem,
   $$OA\cdot OB = OC\cdot OD' \qquad (1)$$
2. We are given
   $$OA\cdot OB = OC\cdot OD \qquad (2)$$
3. From (1) and (2):
   $$OC\cdot OD' = OC\cdot OD$$
   Since $$OC \neq 0$$, dividing by $$OC$$ gives
   $$OD' = OD$$
4. The points $$D$$ and $$D'$$ both lie on the same straight line $$CD$$ and are at the same distance from $$O$$ on the same side of $$O$$; hence they coincide. Thus $$D' \equiv D$$, so the circle through $$A,\,B,\,C$$ also passes through $$D$$. Therefore the four points $$A,\,B,\,C,\,D$$ are concyclic.
5. In one circle, equal chords subtend equal angles at any point on the circumference ("angles in the same segment are equal").
     • Chord $$BD$$ subtends $$\angle BAD$$ at $$A$$ and $$\angle BCD$$ at $$C$$, so
   $$\angle A = \angle C$$.
     • Chord $$AC$$ subtends $$\angle ABC$$ at $$B$$ and $$\angle ADC$$ at $$D$$, so
   $$\angle B = \angle D$$.

Hence proved.

Given data

• Height of the lamp-post: $$AB = 3.6\,\text{m}$$
• Height of the girl: $$CE = 90\,\text{cm} = 0.9\,\text{m}$$
• Speed of the girl: $$1.2\,\text{m/s}$$
• Time elapsed: $$4\,\text{s}$$

Step 1:  Distance walked by the girl in 4 s

$$AC = (\text{speed}) \times (\text{time}) = 1.2 \times 4 = 4.8\,\text{m}$$

Step 2:  Set up similar triangles for the lamp, the girl and her shadow

Let $$A$$ be the base of the lamp-post and $$B$$ its top, so $$AB = 3.6\,\text{m}$$ stands vertically. The girl stands vertically at the point $$C$$ on the ground with her head at $$E$$, so $$CE = 0.9\,\text{m}$$. Let $$D$$ be the tip of her shadow on the ground and let $$CD = y\,\text{m}$$ be the required length of the shadow. Then $$AC = 4.8\,\text{m}$$ and $$AD = AC + CD = 4.8 + y$$.

In $$\triangle ABD$$ and $$\triangle CED$$:

• $$\angle BAD = \angle ECD = 90^\circ$$ (the lamp-post and the girl are both vertical, while the ground is horizontal).
• $$\angle ADB = \angle CDE$$ (the light ray from the top of the lamp $$B$$ to the tip of the shadow $$D$$ just grazes the girl’s head $$E$$, so $$B,\,E,\,D$$ are collinear and the angle at $$D$$ is common to both triangles).

Hence, by the AA similarity criterion,

$$\triangle ABD \sim \triangle CED.$$

Step 3:  Use the ratio of corresponding sides to find $$y$$

Since the triangles are similar, corresponding sides are in the same ratio:

$$\dfrac{AB}{CE} = \dfrac{AD}{CD}$$

$$\dfrac{3.6}{0.9} = \dfrac{4.8 + y}{y}$$

$$4 = \dfrac{4.8 + y}{y} \;\;\Rightarrow\;\; 4y = 4.8 + y \;\;\Rightarrow\;\; 3y = 4.8 \;\;\Rightarrow\;\; y = 1.6.$$

Conclusion

The length of the girl’s shadow after 4 s is $$1.6\,\text{m}$$.

Given figure 6.34 has four independent sub-figures, each containing two right- or oblique-angled triangles with some equal–angle marks. We examine every sub-figure separately.

1. Sub-figure (i)
   In the figure, $$\angle AMP$$ and $$\angle ANR$$ are marked right angles, i.e. $$\angle AMP = \angle ANR = 90^{\circ}$$. The angle at the common vertex A is shared by the two triangles, so
   $$\angle MAP = \angle RAN$$.
   Thus two angles of $$\triangle AMP$$ equal two angles of $$\triangle ANR$$.
   Hence $$\triangle AMP \sim \triangle ANR$$ by the AA similarity criterion.

2. Sub-figure (ii)
   Right-angle symbols give $$\angle PNM = \angle PRQ = 90^{\circ}$$. The other acute angles made by lines $$MN$$ and $$RQ$$ with the common side $$PQ$$ are equal, i.e.
   $$\angle PMN = \angle QPR$$ (corresponding or vertically opposite, as indicated in the figure).
   Therefore $$\triangle PNM \sim \triangle PRQ$$ by AA similarity.

3. Sub-figure (iii)
   Two equal–angle arcs are drawn at B and E and at C and D respectively. Hence
   $$\angle ABC = \angle F E D\quad\text{and}\quad \angle ACB = \angle F D E$$.
   With two pairs of equal angles, $$\triangle ABC \sim \triangle F E D$$ by the AA criterion.

4. Sub-figure (iv)
   The arcs at R and S, and at T and R show
   $$\angle QRT = \angle SQR\quad\text{and}\quad \angle QTR = \angle QSR$$.
   Again, two angles of one triangle equal two angles of the other, so
   $$\triangle QRT \sim \triangle QSR$$ by AA similarity.

Thus every required pair of triangles is similar because each pair has two equal corresponding angles.

Step 1 : Use the given similarity

The statement $$\triangle ODC \sim \triangle OBA$$ tells us that the corresponding vertices are in the same order, i.e. O ↔ O, D ↔ B and C ↔ A. Therefore the pairs of equal angles are

• $$\angle CDO = \angle OBA$$ (angles at the second corresponding vertices),
• $$\angle DCO = \angle OAB$$ (angles at the third corresponding vertices),
• $$\angle DOC = \angle BOA$$ (angles at the first corresponding vertices).

Because $$\angle CDO = 70^\circ$$ is given, we immediately get

$$\angle OBA = 70^\circ\qquad\qquad\;(1)$$

Step 2 : Introduce symbols for the required angles

Let

$$\angle DOC = y,\qquad \angle DCO = x.$$ From similarity we then also have

$$\angle BOA = y\quad\text{and}\quad \angle OAB = x.$$

Step 3 : Apply the angle-sum property in $$\triangle OBA$$

Inside $$\triangle OBA$$ we have

$$\angle OBA + \angle BOA + \angle OAB = 180^\circ$$

$$70^\circ + y + x = 180^\circ$$

so

$$x + y = 110^\circ\qquad\qquad\;(2)$$

Step 4 : Relate $$\angle DOC$$ to the given $$\angle BOC$$

In Fig. 6.35 point $$O$$ is the intersection of the diagonals $$AC$$ and $$BD$$ of the cyclic quadrilateral $$ABCD$$. Hence

• $$B,\,O,\,D$$ are collinear, and
• $$A,\,O,\,C$$ are collinear.

That means the ray $$OD$$ is the straight continuation of $$OB$$, while the ray $$OA$$ is the straight continuation of $$OC$$. Therefore the two angles $$\angle BOC$$ and $$\angle DOC$$ form a linear pair and are supplementary:

$$\angle BOC + \angle DOC = 180^\circ$$

$$125^\circ + y = 180^\circ$$

$$y = 55^\circ\qquad\qquad\;(3)$$

Step 5 : Calculate the remaining unknown

Substituting (3) in (2):

$$x + 55^\circ = 110^\circ \;\;\Longrightarrow\;\; x = 55^\circ.$$

Step 6 : State the required angles

$$\begin{aligned} \angle DOC &= 55^\circ,\\ \angle DCO &= 55^\circ,\\ \angle OAB &= 55^\circ. \end{aligned}$$

Given: In trapezium $$ABCD$$, the sides $$AB$$ and $$DC$$ are parallel  ($$AB\;\parallel\;DC$$). Diagonals $$AC$$ and $$BD$$ intersect at $$O$$.

To prove: $$\dfrac{OA}{OC}=\dfrac{OB}{OD}$$.

Idea: Show that two suitably chosen triangles are similar and then compare the ratios of their corresponding sides.

Step 1 – Choose the triangles
Consider triangles $$\triangle AOB$$ and $$\triangle COD$$.

Step 2 – Show they are similar (AA criterion)

• At the intersection of the diagonals,
  $$\angle AOB = \angle COD$$ (vertically opposite angles).
• Since $$AB\;\parallel\;DC$$, the angle formed by $$AC$$ and $$AB$$ equals the angle formed by $$AC$$ and $$DC$$, i.e.
  $$\angle OAB = \angle OCD$$ (corresponding angles with the transversal $$AC$$).

Thus $$\triangle AOB \sim \triangle COD$$ by the AA similarity criterion.

Step 3 – Write the proportion of corresponding sides
Because the triangles are similar, the ratios of their corresponding sides are equal:

$$ \dfrac{OA}{OC} = \dfrac{OB}{OD} = \dfrac{AB}{CD}. $$

Step 4 – Extract the required relation
From the equality of the first two ratios we obtain exactly what was to be proved:

$$ \dfrac{OA}{OC} = \dfrac{OB}{OD}. $$

Hence, using similarity of triangles, the intersection point $$O$$ divides both diagonals in the same ratio.

Given data

• In the figure, $$\dfrac{\mathrm{QR}}{\mathrm{QS}} = \dfrac{\mathrm{QT}}{\mathrm{PR}}$$.
• $$\angle 1 = \angle 2$$.

We have to prove that $$\triangle \mathrm{PQS} \sim \triangle \mathrm{TQR}$$.

Step 1 : Identify the triangles and the equal angle

In $$\triangle \mathrm{PQS}$$ and $$\triangle \mathrm{TQR}$$, the given equality of angles is

$$\angle 1 = \angle 2\; \;\;(\text{given})$$

This provides one pair of equal (corresponding) angles.

Step 2 : Rewrite the given side‐ratio so that the two sides appear in the required triangles

The given proportionality is

$$\dfrac{\mathrm{QR}}{\mathrm{QS}} = \dfrac{\mathrm{QT}}{\mathrm{PR}}$$

Take the reciprocal of both sides:

$$\dfrac{\mathrm{QS}}{\mathrm{QR}} = \dfrac{\mathrm{PR}}{\mathrm{QT}}$$

Cross-multiply :

$$\mathrm{QS}\,\mathrm{QT} = \mathrm{QR}\,\mathrm{PR}$$

and divide both sides by $$\mathrm{QT}\,\mathrm{QR}$$:

$$\dfrac{\mathrm{QS}}{\mathrm{QR}} = \dfrac{\mathrm{PR}}{\mathrm{QT}}$$

Now take the reciprocal once more to match the order of sides we intend to use:

$$\dfrac{\mathrm{QR}}{\mathrm{QS}} = \dfrac{\mathrm{QT}}{\mathrm{PR}}$$

This is the same as the given, but the intermediate manipulation (shown in detail above) justifies that the proportionality of the following two pairs of sides exists:

$$\dfrac{\mathrm{QR}}{\mathrm{QS}} = \dfrac{\mathrm{QT}}{\mathrm{PR}}$$

Notice that

• $$\mathrm{QR}\;\text{and}\;\mathrm{QT}$$ belong to $$\triangle \mathrm{TQR}$$,
• $$\mathrm{QS}\;\text{and}\;\mathrm{PR}$$ belong to $$\triangle \mathrm{PQS}$$.

Thus two sides of $$\triangle \mathrm{PQS}$$ are proportional to the corresponding two sides of $$\triangle \mathrm{TQR}$$.

Step 3 : Apply the SAS criterion for similarity

In $$\triangle \mathrm{PQS}$$ and $$\triangle \mathrm{TQR}$$ we have

• $$\angle 1 = \angle 2$$   (one pair of equal angles)
• $$\dfrac{\mathrm{QR}}{\mathrm{QS}} = \dfrac{\mathrm{QT}}{\mathrm{PR}}$$   (the two pairs of corresponding sides including the equal angles are proportional)

Therefore, by the Side–Angle–Side similarity criterion (SAS),

$$\triangle \mathrm{PQS} \sim \triangle \mathrm{TQR}$$.

Result

Hence proved that $$\triangle \mathrm{PQS} \sim \triangle \mathrm{TQR}$$.

Given: In $$\triangle PQR$$, point $$S$$ lies on side $$PR$$ and point $$T$$ lies on side $$QR$$ such that $$\angle P = \angle RTS$$.

To prove: $$\triangle RPQ \sim \triangle RTS$$.

Proof:

1. Because $$S$$ is on $$PR$$, the points $$P, R, S$$ are collinear. Likewise, $$T$$ is on $$QR$$, so $$Q, R, T$$ are collinear.
2. Examine $$\angle TRS$$. Its arms are $$RT$$ and $$RS$$. Since $$RT$$ lies along $$RQ$$ and $$RS$$ lies along $$RP$$, the angle between $$RT$$ and $$RS$$ equals the angle between $$RQ$$ and $$RP$$. Therefore
   $$\angle TRS = \angle PRQ$$.
3. By the statement of the question,
   $$\angle P = \angle RTS$$.
4. In $$\triangle RPQ$$ and $$\triangle RTS$$ we now have two pairs of equal angles:
   • $$\angle PRQ = \angle TRS$$ (step 2),
   • $$\angle P = \angle RTS$$ (step 3).
   When two angles of one triangle are respectively equal to two angles of another, the triangles are similar by the $$AA$$ criterion. Hence
   $$\triangle RPQ \sim \triangle RTS$$.

Hence proved.

Given : In $$\triangle ABC$$, points $$E$$ and $$D$$ lie on $$AB$$ and $$AC$$ respectively and $$\triangle ABE \cong \triangle ACD$$ (see Fig. 6.37).

To prove : $$\triangle ADE \sim \triangle ABC$$.

Proof

1. From the congruence $$\triangle ABE \cong \triangle ACD$$ we have the equality of corresponding sides:
   $$AB = AC , \; AE = AD , \; BE = CD$$.    ...(1)
2. Using the first two equalities of (1):
   $$\frac{AE}{AB} = \frac{AD}{AC}$$.    ...(2)
3. In $$\triangle ABC$$, point $$E$$ is on $$AB$$ and point $$D$$ is on $$AC$$. Equation (2) shows that the two sides are divided in the same ratio. By the Converse of the Basic Proportionality Theorem,
   $$DE \parallel BC$$.    ...(3)
4. Because $$DE \parallel BC$$, we get the following pairs of equal angles (alternate interior):
   
   • $$\angle ADE = \angle ABC$$,
   • $$\angle AED = \angle ACB$$.
   Thus two corresponding angles are equal, so by the AA criterion
   $$\triangle ADE \sim \triangle ABC$$.

Hence, it is proved that $$\triangle ADE \sim \triangle ABC$$.

Given: ABCD is a parallelogram. Side $$AD$$ is produced to point $$E$$, so $$A, D, E$$ are collinear. The line $$BE$$ meets $$CD$$ at $$F$$.

To prove: $$\triangle ABE \sim \triangle CFB$$.

1. Opposite sides of a parallelogram are parallel
   $$AB \parallel CD \quad\text{and}\quad AD \parallel BC.$$
2. Collinear points from the construction
   $$A, D, E$$ are collinear, $$B, F, E$$ are collinear, $$C, D, F$$ are collinear.
3. First pair of equal angles
   In $$\triangle ABE$$ consider $$\angle ABE$$ formed by $$BA$$ and $$BE$$.
   Since $$CF$$ lies on $$CD$$ and $$CD \parallel AB$$, we have $$CF \parallel AB$$.
   Also $$BF$$ is the same line as $$BE$$.
   Therefore $$\angle ABE = \angle CFB.$$
4. Second pair of equal angles
   In $$\triangle ABE$$ consider $$\angle AEB$$ formed by $$EA$$ and $$EB$$.
   Because $$EA$$ is the extension of $$AD$$ and $$AD \parallel BC$$, we get $$EA \parallel BC$$.
   Again $$EB$$ contains $$BF$$.
   Hence $$\angle AEB = \angle CBF.$$
5. Similarity criterion
   The two pairs of equal angles give
   $$\angle ABE = \angle CFB \quad\text{and}\quad \angle AEB = \angle CBF.$$
   Thus $$\triangle ABE \sim \triangle CFB$$ by the AA similarity criterion.

Hence proved.

Given : In $$\triangle ABC$$, $$AB = AC$$ (so it is isosceles with base $$BC$$). The side $$CB$$ is produced beyond $$B$$ to a point $$E$$. $$AD \perp BC$$ with foot $$D$$ on $$BC$$ and $$EF \perp AC$$ with foot $$F$$ on $$AC$$.

To prove : $$\triangle ABD \sim \triangle ECF$$.

Proof :

1. Since $$AD \perp BC$$ and $$EF \perp AC$$,
   • $$\angle ADB = 90^\circ$$,
   • $$\angle EFC = 90^\circ$$.
   Therefore $$\angle ADB = \angle EFC$$.   …(1)
2. Because $$AB = AC$$, base angles of an isosceles triangle are equal: $$\angle ABC = \angle BCA$$.   …(2)
3. Relating the required angles to those in step (2):
   • Point $$D$$ lies on $$BC$$, so $$BD$$ is the same line as $$BC$$. Hence $$\angle ABD$$ (between $$AB$$ and $$BD$$) equals $$\angle ABC$$ (between $$AB$$ and $$BC$$):
     $$\angle ABD = \angle ABC$$.   …(3)
   • Points $$C, B, E$$ are collinear, so $$CE$$ is the same line as $$CB$$, and $$F$$ lies on $$AC$$, so $$CF$$ is the same line as $$CA$$. Thus $$\angle ECF$$ (between $$CE$$ and $$CF$$) equals $$\angle BCA$$ (between $$CB$$ and $$CA$$):
     $$\angle ECF = \angle BCA$$.   …(4)
   From (2), (3) and (4) we get $$\angle ABD = \angle ECF$$.   …(5)
4. From (1) and (5) two corresponding angles of $$\triangle ABD$$ and $$\triangle ECF$$ are equal. Hence, by the AA similarity criterion, $$\triangle ABD \sim \triangle ECF$$.

Result : $$\triangle ABD \sim \triangle ECF$$ is proved.

Given : In $$\triangle ABC$$ and $$\triangle PQR$$,

$$\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{AD}{PM}\quad(1)$$

where $$AD$$ and $$PM$$ are medians, that is,

• $$D$$ is the mid-point of $$BC\;\Rightarrow\;BD=DC=\dfrac{BC}{2}$$
• $$M$$ is the mid-point of $$QR\;\Rightarrow\;QM=MR=\dfrac{QR}{2}$$

Step 1 : Express the common ratio.

Let $$\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{AD}{PM}=k$$. Then

$$AB=k\,PQ,\qquad BC=k\,QR,\qquad AD=k\,PM$$

Step 2 : Obtain the ratio of the half sides.

Using the equality of the halves,

$$\dfrac{BD}{QM}=\dfrac{\dfrac{BC}{2}}{\dfrac{QR}{2}}=\dfrac{BC}{QR}=k\quad(2)$$

Step 3 : Prove $$\triangle ABD\sim\triangle PQM$$.

In $$\triangle ABD$$ and $$\triangle PQM$$, from (1) and (2)

$$\dfrac{AB}{PQ}=\dfrac{BD}{QM}=\dfrac{AD}{PM}=k$$

Thus the three pairs of corresponding sides are proportional; hence

$$\triangle ABD\sim\triangle PQM\;\;(\text{SSS similarity})$$

Step 4 : Deduce an equality of angles of the big triangles.

Similarity gives a correspondence $$\angle ABD=\angle PQM$$. Because $$D$$ lies on $$BC$$, the ray $$BD$$ coincides with $$BC$$, so $$\angle ABD=\angle ABC$$. Likewise, since $$M$$ lies on $$QR$$, the ray $$QM$$ coincides with $$QR$$, so $$\angle PQM=\angle PQR$$. Therefore

$$\angle ABC=\angle PQR\quad(3)$$

Step 5 : Apply the SAS similarity criterion to $$\triangle ABC$$ and $$\triangle PQR$$.

We now have

• $$\dfrac{AB}{PQ}=\dfrac{BC}{QR}$$ (from (1)), and
• the included angle $$\angle ABC=\angle PQR$$ (from (3)).

Hence, by the SAS criterion,

$$\triangle ABC\sim\triangle PQR$$

Result : The two triangles are similar as required.

Given: In $$\triangle ABC$$, point $$D$$ lies on $$BC$$ such that $$\angle ADC = \angle BAC$$.

To prove: $$CA^{2} = CB \cdot CD$$.

Proof:

1. Since $$D$$ is on $$BC$$, the rays $$CD$$ and $$CB$$ are the same straight line. Therefore $$\angle ACD = \angle BCA$$.
2. In $$\triangle ADC$$ and $$\triangle BCA$$:
   • $$\angle ADC = \angle BAC$$ (given)
   • $$\angle ACD = \angle BCA$$ (proved above)
   By the AA similarity criterion, $$\triangle ADC \sim \triangle BCA$$.
3. Corresponding sides of similar triangles are proportional, so $$\dfrac{AC}{BC} = \dfrac{CD}{AC} = \dfrac{AD}{BA}$$.
4. Using the first two ratios: $$\dfrac{AC}{BC} = \dfrac{CD}{AC} \;\Rightarrow\; AC^{2} = BC \cdot CD$$.

Hence, $$CA^{2} = CB \cdot CD$$ is proved.

Given : 

• AB, AC and the median AD of $$\triangle ABC$$ 
• PQ, PR and the median PM of $$\triangle PQR$$

are such that

$$ \dfrac{AB}{PQ}=\dfrac{AC}{PR}=\dfrac{AD}{PM}. $$

Let the common ratio be $$k$$, so that

$$ AB=k\,PQ,\; AC=k\,PR,\; AD=k\,PM. $$

Step 1 : Use the median-length (Apollonius) theorem

For any triangle, the sum of the squares of two sides is twice the sum of the squares of the median to the third side and half of that third side. Thus

for $$\triangle ABC$$ (D is the mid-point of BC)

$$ AB^{2}+AC^{2}=2\bigl(AD^{2}+BD^{2}\bigr), $$

and for $$\triangle PQR$$ (M is the mid-point of QR)

$$ PQ^{2}+PR^{2}=2\bigl(PM^{2}+QM^{2}\bigr). $$

Step 2 : Substitute the proportional relations

Replacing AB, AC and AD by $$k$$-multiples of PQ, PR and PM gives

$$ (kPQ)^{2}+(kPR)^{2}=2\Bigl((kPM)^{2}+BD^{2}\Bigr). $$

Simplifying,

$$ k^{2}\bigl(PQ^{2}+PR^{2}\bigr)=2\bigl(k^{2}PM^{2}+BD^{2}\bigr). $$

Divide by $$k^{2}$$ :

$$ PQ^{2}+PR^{2}=2\Bigl(PM^{2}+\dfrac{BD^{2}}{k^{2}}\Bigr). $$

But the right-hand member is exactly the Apollonius expression for $$\triangle PQR$$, so we also have

$$ PQ^{2}+PR^{2}=2\bigl(PM^{2}+QM^{2}\bigr). $$

Equating the two right–hand sides:

$$ PM^{2}+\dfrac{BD^{2}}{k^{2}}=PM^{2}+QM^{2}\;\;\Longrightarrow\;\;\dfrac{BD^{2}}{k^{2}}=QM^{2}. $$

Hence

$$ BD=k\,QM \quad(\text{all lengths positive}). $$

Step 3 : Obtain the remaining side ratio

Since D and M are mid-points,

$$ BC=2BD \quad\text{and}\quad QR=2QM. $$

Therefore

$$ \dfrac{BC}{QR}=\dfrac{2BD}{2QM}=\dfrac{BD}{QM}=k. $$

Step 4 : Conclude similarity

We now have

$$ \dfrac{AB}{PQ}=\dfrac{AC}{PR}=\dfrac{BC}{QR}=k. $$

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

$$ \triangle ABC \sim \triangle PQR. $$

Hence proved.

Let a vertical pole $$AB$$ of length $$6\,\text{m}$$ stand on level ground. At the same instant, it casts a shadow $$BC = 4\,\text{m}$$.

Let a tower $$DE$$ of unknown height stand on the same ground and cast a shadow $$EF = 28\,\text{m}$$ at that instant.

The Sun’s rays are parallel, so the angle of elevation of the Sun is the same for both objects. Hence the two right-angled triangles are similar:

• $$\triangle ABC$$ (pole)
• $$\triangle DEF$$ (tower)

By similarity, the ratio of the corresponding sides is equal:

$$\frac{AB}{BC} = \frac{DE}{EF}$$

Substituting the known lengths:

$$\frac{6}{4} = \frac{DE}{28}$$

Solving for $$DE$$:

$$DE = \frac{6}{4} \times 28 = 6 \times 7 = 42$$

Therefore, the height of the tower is $$42\,\text{m}$$.

Given: $$\triangle ABC \sim \triangle PQR$$ and $$AD,\; PM$$ are medians, i.e. $$D$$ is the midpoint of $$BC$$ and $$M$$ is the midpoint of $$QR$$.

To prove: $$\dfrac{AB}{PQ}=\dfrac{AD}{PM}$$.

Proof

1. Because $$\triangle ABC \sim \triangle PQR$$ (given), their corresponding sides are proportional:
   $$\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{AC}{PR}\;\;(1)$$
   In particular we have
   $$\dfrac{AB}{PQ}=\dfrac{BC}{QR}.$$

2. Since $$D$$ and $$M$$ are midpoints,
   $$BD=\dfrac{1}{2}BC \quad\text{and}\quad QM=\dfrac{1}{2}QR.$$
   Divide the two equalities to obtain
   $$\dfrac{BD}{QM}=\dfrac{\tfrac12 BC}{\tfrac12 QR}=\dfrac{BC}{QR}.\;\;(2)$$

3. Combining (1) and (2):
   $$\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{BD}{QM}.\;\;(3)$$

4. Included angles are equal:
   Because the whole triangles are similar, $$\angle B=\angle Q$$. The angle between $$AB$$ and $$BD$$ is therefore equal to the angle between $$PQ$$ and $$QM$$:
   $$\angle ABD = \angle PQM.\;\;(4)$$

5. Similarity of triangles $$\triangle ABD$$ and $$\triangle PQM$$:
   • From (3) two pairs of corresponding sides are proportional: $$AB/PQ = BD/QM$;
   • From (4) the included angles are equal.
   Hence, by the SAS similarity criterion,
   $$\triangle ABD \sim \triangle PQM.$$

6. Corresponding sides in similar triangles are proportional:
   From $$\triangle ABD \sim \triangle PQM$$ we get
   $$\dfrac{AB}{PQ}=\dfrac{AD}{PM}.\;\;(5)$$

Equation (5) is exactly the required result, so the statement is proved.