NCERT Solutions For Class 10 Maths: Triangles

EXERCISE 6.1

Question 1: Fill in the blanks using the correct word given in brackets :
(i) All circles are . (congruent, similar)
(ii) All squares are . (similar, congruent)
(iii) All triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are and (b) their corresponding sides are . (equal, proportional)

Solution-

1.(i) similar
1.(ii) similar
1.(iii) equilateral
1.(iv) (a) equal, (b) proportional

Question 2: Give two different examples of pair of
(i) similar figures. (ii) non-similar figures.

Solution-

2.(i) Two different examples of pairs of similar figures are:
1.All circles.
2.All equilateral triangles.

2.(ii) Two different examples of pairs of non-similar figures are:
1.A circle and a square.
2.A square and a rhombus (since corresponding angles are not equal).

Question 3: State whether the following quadrilaterals are similar or not:

Solution-

The quadrilaterals (a square and a rhombus, as depicted in Fig. 6.7/6.8 referenced in the source) are not similar.
The corresponding sides may be in the same ratio (proportional), but their corresponding angles are not equal. Both conditions—corresponding angles being equal and corresponding sides being proportional—must be satisfied for polygons to be similar.

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EXERCISE 6.2

Question 1: In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution-

1.Applying the Basic Proportionality Theorem (Theorem 6.1), $\frac{AD}{DB} = \frac{AE}{EC}$.

1.(i) Given $AD = 1.5 \text{ cm}, DB = 3 \text{ cm}, AE = 1 \text{ cm}$.
$\frac{1.5}{3} = \frac{1}{EC}$
$EC = \frac{3 \times 1}{1.5} = 2 \text{ cm}$.

1.(ii) Given $DB = 7.2 \text{ cm}, AE = 1.8 \text{ cm}, EC = 5.4 \text{ cm}$.
$\frac{AD}{7.2} = \frac{1.8}{5.4}$
$AD = \frac{1.8}{5.4} \times 7.2 = \frac{1}{3} \times 7.2 = 2.4 \text{ cm}$.

Question 2: E and F are points on the sides PQ and PR respectively of a $\triangle PQR$. For each of the following cases, state whether EF || QR :
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

Solution-

2.(i) Check ratios: $\frac{PE}{EQ} = \frac{3.9}{3} = 1.3$.
$\frac{PF}{FR} = \frac{3.6}{2.4} = 1.5$.
Since $\frac{PE}{EQ} \neq \frac{PF}{FR}$, EF is NOT parallel to QR.

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

Solution-

2.(ii) Check ratios: $\frac{PE}{QE} = \frac{4}{4.5} = \frac{8}{9}$.
$\frac{PF}{RF} = \frac{8}{9}$.
Since $\frac{PE}{QE} = \frac{PF}{RF}$, EF is parallel to QR (by Theorem 6.2).

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Solution-

2.(iii) Calculate remaining segments: $EQ = PQ - PE = 1.28 - 0.18 = 1.10 \text{ cm}$.
$FR = PR - PF = 2.56 - 0.36 = 2.20 \text{ cm}$.
Check ratios: $\frac{PE}{EQ} = \frac{0.18}{1.10} = \frac{18}{110} = \frac{9}{55}$.
$\frac{PF}{FR} = \frac{0.36}{2.20} = \frac{36}{220} = \frac{9}{55}$.
Since $\frac{PE}{EQ} = \frac{PF}{FR}$, EF is parallel to QR (by Theorem 6.2).

Question 3: In Fig. 6.18, if LM || CB and LN || CD, prove that $\frac{AM}{AB} = \frac{AN}{AD}$

Solution-

3.1. In $\triangle ABC$, $LM || CB$. By Theorem 6.1, $\frac{AM}{MB} = \frac{AL}{LC}$.
3.2. In $\triangle ADC$, $LN || CD$. By Theorem 6.1, $\frac{AN}{ND} = \frac{AL}{LC}$.
3.3. From steps 3.1 and 3.2, $\frac{AM}{MB} = \frac{AN}{ND}$.
3.4. Taking reciprocals and adding 1: $\frac{MB}{AM} + 1 = \frac{ND}{AN} + 1$.
3.5. $\frac{MB + AM}{AM} = \frac{ND + AN}{AN}$. Since $MB + AM = AB$ and $ND + AN = AD$.
3.6. $\frac{AB}{AM} = \frac{AD}{AN}$. Taking reciprocals again, $\frac{AM}{AB} = \frac{AN}{AD}$.

Question 4: In Fig. 6.19, DE || AC and DF || AE. Prove that $\frac{BF}{FE} = \frac{BE}{EC}$

Solution-

4.1. In $\triangle ABE$, $DF || AE$. By Theorem 6.1 (BPT), $\frac{BD}{DA} = \frac{BF}{FE}$.
4.2. In $\triangle ABC$, $DE || AC$. By Theorem 6.1 (BPT), $\frac{BD}{DA} = \frac{BE}{EC}$.
4.3. Since both ratios are equal to $\frac{BD}{DA}$, $\frac{BF}{FE} = \frac{BE}{EC}$.

Question 5: In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.

Solution-

5.1. In $\triangle POQ$, $DE || OQ$. By Theorem 6.1 (BPT), $\frac{PE}{EQ} = \frac{PD}{DO}$.
5.2. In $\triangle POR$, $DF || OR$. By Theorem 6.1 (BPT), $\frac{PF}{FR} = \frac{PD}{DO}$.
5.3. From steps 5.1 and 5.2, $\frac{PE}{EQ} = \frac{PF}{FR}$.
5.4. In $\triangle PQR$, since $E$ and $F$ divide sides $PQ$ and $PR$ in the same ratio, by Theorem 6.2 (Converse of BPT), $EF || QR$.

Question 6: In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Solution-

6.1. In $\triangle OPQ$, $AB || PQ$. By Theorem 6.1 (BPT), $\frac{OA}{AP} = \frac{OB}{BQ}$.
6.2. In $\triangle OPR$, $AC || PR$. By Theorem 6.1 (BPT), $\frac{OA}{AP} = \frac{OC}{CR}$.
6.3. From steps 6.1 and 6.2, $\frac{OB}{BQ} = \frac{OC}{CR}$.
6.4. In $\triangle OQR$, since $B$ and $C$ divide sides $OQ$ and $OR$ in the same ratio, by Theorem 6.2 (Converse of BPT), $BC || QR$.

Question 7: Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Solution-

7.1. Let $\triangle ABC$ have $D$ as the mid-point of $AB$, so $\frac{AD}{DB} = 1$. Let $DE$ be a line segment such that $DE || BC$ and $E$ is on $AC$.
7.2. By Theorem 6.1 (BPT), since $DE || BC$, $\frac{AD}{DB} = \frac{AE}{EC}$.
7.3. Since $D$ is the mid-point, $\frac{AD}{DB} = 1$.
7.4. Therefore, $1 = \frac{AE}{EC}$, which implies $AE = EC$.
7.5. Thus, $E$ is the mid-point of $AC$, meaning the third side is bisected.

Question 8: Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Solution-

8.1. Let $\triangle ABC$ have $D$ and $E$ as the mid-points of sides $AB$ and $AC$ respectively.
8.2. Since $D$ is the mid-point of $AB$, $AD = DB$, so $\frac{AD}{DB} = 1$.
8.3. Since $E$ is the mid-point of $AC$, $AE = EC$, so $\frac{AE}{EC} = 1$.
8.4. Therefore, $\frac{AD}{DB} = \frac{AE}{EC}$.
8.5. By Theorem 6.2 (Converse of BPT), since the line $DE$ divides two sides of the triangle in the same ratio, $DE$ is parallel to the third side $BC$.

Question 9: ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that $\frac{AO}{BO} = \frac{CO}{DO}$.

Solution-

9.1. Given $AB || DC$. Draw a line $EO$ through $O$ parallel to $AB$ (and thus $DC$), where $E$ is on $AD$.
9.2. In $\triangle DAB$, $EO || AB$. By Theorem 6.1, $\frac{DE}{EA} = \frac{DO}{OB}$.
9.3. In $\triangle ADC$, $EO || DC$. By Theorem 6.1, $\frac{AE}{ED} = \frac{AO}{CO}$.
9.4. From step 9.2, $\frac{EA}{DE} = \frac{OB}{DO}$.
9.5. Equating the ratio $\frac{EA}{DE}$ from step 9.4 and $\frac{CO}{AO}$ from step 9.3: $\frac{OB}{DO} = \frac{CO}{AO}$.
9.6. Rearranging to the required form: $\frac{AO}{BO} = \frac{CO}{DO}$.

Question 10: The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\frac{AO}{BO} = \frac{CO}{DO}$. Show that ABCD is a trapezium.

Solution-

10.1. Given $\frac{AO}{BO} = \frac{CO}{DO}$. Rearrange: $\frac{AO}{CO} = \frac{BO}{DO}$.
10.2. Draw $EO$ through $O$ parallel to $AB$, with $E$ on $AD$.
10.3. In $\triangle DAB$, $EO || AB$. By Theorem 6.1, $\frac{DE}{EA} = \frac{DO}{OB}$.
10.4. From the given ratio, $\frac{DO}{OB} = \frac{CO}{AO}$ (by taking reciprocals of the rearranged ratio).
10.5. From steps 10.3 and 10.4, $\frac{DE}{EA} = \frac{CO}{AO}$. Taking reciprocals: $\frac{EA}{DE} = \frac{AO}{CO}$.
10.6. In $\triangle ADC$, the line $EO$ divides $AD$ and $AC$ such that $\frac{AE}{ED} = \frac{AO}{OC}$.
10.7. By Theorem 6.2 (Converse of BPT), $EO || DC$.
10.8. Since $EO || AB$ (by construction) and $EO || DC$, it follows that $AB || DC$.
10.9. Therefore, $ABCD$ is a trapezium.

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EXERCISE 6.3

Question 1: State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :

Solution-

1.(i) $\triangle ABC$ and $\triangle PQR$: $\angle A = \angle P = 60^{\circ}$, $\angle B = \angle Q = 80^{\circ}$, $\angle C = \angle R = 40^{\circ}$.
Similar. Criterion: AAA similarity criterion. Symbolic form: $\triangle ABC \sim \triangle PQR$.

1.(ii) $\triangle ABC$ and $\triangle QRP$: $\frac{AB}{QR} = \frac{2}{4} = \frac{1}{2}$; $\frac{BC}{RP} = \frac{2.5}{5} = \frac{1}{2}$; $\frac{CA}{PQ} = \frac{3}{6} = \frac{1}{2}$.
Similar. Criterion: SSS similarity criterion. Symbolic form: $\triangle ABC \sim \triangle QRP$.

1.(iii) $\triangle LMP$ and $\triangle DEF$: Ratios are $\frac{2}{4} = \frac{1}{2}$, $\frac{3}{6} = \frac{1}{2}$, $\frac{2.7}{5}$. The sides are not in the same ratio.
Not similar.

1.(iv) $\triangle MNL$ and $\triangle QPR$: $\angle M = \angle Q = 70^{\circ}$.
$\frac{MN}{PQ} = \frac{2.5}{5} = \frac{1}{2}$; $\frac{ML}{QR} = \frac{5}{10} = \frac{1}{2}$.
Similar. Criterion: SAS similarity criterion. Symbolic form: $\triangle NML \sim \triangle PQR$.

1.(v) $\triangle ABC$ and $\triangle DEF$: $\angle A = 80^{\circ}, \angle F = 80^{\circ}$. Sides $\frac{2.5}{3} \neq \frac{5}{6}$. The sides including the equal angles are not proportional, and the given sides are not proportional in any order.
Not similar.

1.(vi) $\triangle DEF$ and $\triangle PQR$: Calculate third angles. $\angle F = 180^{\circ} - 70^{\circ} - 80^{\circ} = 30^{\circ}$. $\angle P = 180^{\circ} - 80^{\circ} - 30^{\circ} = 70^{\circ}$.
$\angle D = \angle P = 70^{\circ}$, $\angle E = \angle Q = 80^{\circ}$, $\angle F = \angle R = 30^{\circ}$.
Similar. Criterion: AAA (or AA) similarity criterion. Symbolic form: $\triangle DEF \sim \triangle PQR$.

Question 2: In Fig. 6.35, $\triangle ODC \sim \triangle OBA$, $\angle BOC = 125^{\circ}$ and $\angle CDO = 70^{\circ}$. Find $\angle DOC, \angle DCO$ and $\angle OAB$.

Solution-

2.1. $\angle DOC = 180^{\circ} - \angle BOC = 180^{\circ} - 125^{\circ} = 55^{\circ}$ (Linear pair).
2.2. In $\triangle ODC$, $\angle DCO = 180^{\circ} - \angle CDO - \angle DOC = 180^{\circ} - 70^{\circ} - 55^{\circ} = 55^{\circ}$ (Angle sum property).
2.3. Since $\triangle ODC \sim \triangle OBA$, corresponding angles are equal.
$\angle OAB = \angle OCD = 55^{\circ}$. (The correspondence is $O \leftrightarrow O, D \leftrightarrow B, C \leftrightarrow A$).
Wait, the given similarity is $\triangle ODC \sim \triangle OBA$. The correspondence is $O \leftrightarrow O, D \leftrightarrow B, C \leftrightarrow A$.
Thus, $\angle OAB = \angle OCD$ is incorrect. $\angle OAB$ corresponds to $\angle ODC$ or $\angle A$ corresponds to $\angle C$.
From the similarity: $\angle OAB$ corresponds to $\angle ODC$.
$\angle OAB = \angle ODC = 70^{\circ}$.

$\angle DOC = 55^{\circ}$.
$\angle DCO = 55^{\circ}$.
$\angle OAB = 70^{\circ}$.

Question 3: Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that $\frac{OA}{OC} = \frac{OB}{OD}$.

Solution-

3.1. Consider $\triangle AOB$ and $\triangle COD$.
3.2. $\angle AOB = \angle COD$ (Vertically opposite angles).
3.3. Since $AB || DC$, $\angle OAB = \angle OCD$ (Alternate angles).
3.4. By AA similarity criterion, $\triangle AOB \sim \triangle COD$.
3.5. Since the triangles are similar, the ratio of corresponding sides is equal: $\frac{AO}{CO} = \frac{BO}{DO}$.
3.6. Rearranging the proportion: $\frac{OA}{OB} = \frac{OC}{OD}$. (The required form is $\frac{OA}{OC} = \frac{OB}{OD}$, which is proven directly by $\frac{AO}{CO} = \frac{BO}{DO}$).

Question 4: In Fig. 6.36, $\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$. Show that $\triangle PQS \sim \triangle TQR$.

Solution-

4.1. In $\triangle PQR$, since $\angle 1 = \angle 2$, the sides opposite to these angles are equal: $PR = PQ$.
4.2. Substitute $PQ$ for $PR$ in the given proportion: $\frac{QR}{QS} = \frac{QT}{PQ}$.
4.3. Rearrange the terms: $\frac{PQ}{QS} = \frac{QT}{QR}$.
4.4. Consider $\triangle PQS$ and $\triangle TQR$:
$\angle PQS = \angle TQR$ ($\angle 1$ is common).
The sides including the common angle are proportional: $\frac{PQ}{QS} = \frac{QT}{QR}$.
4.5. By the SAS similarity criterion, $\triangle PQS \sim \triangle TQR$.

Question 5: S and T are points on sides PR and QR of $\triangle PQR$ such that $\angle P = \angle RTS$. Show that $\triangle RPQ \sim \triangle RTS$.

Solution-

5.1. Consider $\triangle RPQ$ and $\triangle RTS$.
5.2. $\angle P = \angle RTS$ (Given).
5.3. $\angle PRQ = \angle TRS$ (Angle $\angle R$ is common to both triangles).
5.4. By the AA similarity criterion, $\triangle RPQ \sim \triangle RTS$.

Question 6: In Fig. 6.37, if $\triangle ABE \cong \triangle ACD$, show that $\triangle ADE \sim \triangle ABC$.

Solution-

6.1. Given $\triangle ABE \cong \triangle ACD$. By CPCT, $AB = AC$ and $AE = AD$.
6.2. From $AB = AC$, we have $\frac{AB}{AC} = 1$.
6.3. From $AE = AD$, we have $\frac{AE}{AD} = 1$.
6.4. Dividing $AE = AD$ by $AB = AC$: $\frac{AE}{AB} = \frac{AD}{AC}$. Rearrange this as $\frac{AD}{AB} = \frac{AE}{AC}$.
6.5. Consider $\triangle ADE$ and $\triangle ABC$.
The sides are proportional: $\frac{AD}{AB} = \frac{AE}{AC}$.
The included angle is common: $\angle DAE = \angle BAC$ ($\angle A$ is common).
6.6. By the SAS similarity criterion, $\triangle ADE \sim \triangle ABC$.

Question 7: In Fig. 6.38, altitudes AD and CE of $\triangle ABC$ intersect each other at the point P. Show that:
(i) $\triangle AEP \sim \triangle CDP$

Solution-

7.(i) 1. In $\triangle AEP$ and $\triangle CDP$: $\angle AEP = 90^{\circ}$ ($CE \perp AB$) and $\angle CDP = 90^{\circ}$ ($AD \perp BC$). No, the right angles are $\angle AEP$ and $\angle CDP$.
2. $\angle AEP = 90^{\circ}$ is not correct based on altitude definition. The right angle is $\angle AEP$ and $\angle CDP$. Wait, $\angle ADC=90^{\circ}$ and $\angle AEC=90^{\circ}$.
3. $\angle AEP = 90^{\circ}$ (since $CE$ is altitude to $AB$).
4. $\angle APE = \angle CPD$ (Vertically opposite angles).
5. By AA similarity criterion, $\triangle AEP \sim \triangle CDP$.

(ii) $\triangle ABD \sim \triangle CBE$

Solution-

7.(ii) 1. In $\triangle ABD$ and $\triangle CBE$: $\angle ADB = 90^{\circ}$ ($AD \perp BC$) and $\angle CEB = 90^{\circ}$ ($CE \perp AB$).
2. $\angle ABD = \angle CBE$ (Angle $\angle B$ is common).
3. By AA similarity criterion, $\triangle ABD \sim \triangle CBE$.

(iii) $\triangle AEP \sim \triangle ADB$

Solution-

7.(iii) 1. In $\triangle AEP$ and $\triangle ADB$: $\angle PAE = \angle DAB$ (Angle $\angle A$ is common).
2. $\angle AEP = 90^{\circ}$ ($CE \perp AB$) and $\angle ADB = 90^{\circ}$ ($AD \perp BC$).
3. By AA similarity criterion, $\triangle AEP \sim \triangle ADB$. (Note: $\angle AEP$ corresponds to $\angle ADB$ in the similarity criteria based on right angle and common angle $A$).

(iv) $\triangle PDC \sim \triangle BEC$

Solution-

7.(iv) 1. In $\triangle PDC$ and $\triangle BEC$: $\angle PDC = 90^{\circ}$ ($AD \perp BC$) and $\angle BEC = 90^{\circ}$ ($CE \perp AB$).
2. $\angle PCD = \angle BCE$ (Angle $\angle C$ is common).
3. By AA similarity criterion, $\triangle PDC \sim \triangle BEC$.

Question 8: E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that $\triangle ABE \sim \triangle CFB$.

Solution-

8.1. Given $ABCD$ is a parallelogram, so $AE || BC$ (as $AD || BC$).
8.2. Consider $\triangle ABE$ and $\triangle CFB$.
8.3. $\angle AEB = \angle CBF$ (Alternate interior angles, since $AE || BC$ and $BE$ is the transversal).
8.4. $\angle BAE = \angle FCE$ is incorrect. $\angle A = \angle C$.
8.5. $\angle BAE = \angle BCF$ (Opposite angles of a parallelogram are equal, i.e., $\angle A = \angle C$).
8.6. By the AA similarity criterion, $\triangle ABE \sim \triangle CFB$.

Question 9: In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) $\triangle ABC \sim \triangle AMP$

Solution-

9.(i) 1. In $\triangle ABC$ and $\triangle AMP$: $\angle ABC = 90^{\circ}$ and $\angle AMP = 90^{\circ}$. So, $\angle ABC = \angle AMP$.
2. $\angle BAC = \angle MAP$ (Angle $\angle A$ is common).
3. By the AA similarity criterion, $\triangle ABC \sim \triangle AMP$.

(ii) $\frac{CA}{PA} = \frac{BC}{MP}$

Solution-

9.(ii) Since $\triangle ABC \sim \triangle AMP$ (from part i), the corresponding sides are proportional: $\frac{CA}{PA} = \frac{BC}{MP}$ (Corresponding sides of similar triangles are proportional).

Question 10: CD and GH are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that D and H lie on sides AB and FE of $\triangle ABC$ and $\triangle EFG$ respectively. If $\triangle ABC \sim \triangle FEG$, show that:
(i) $\frac{CD}{GH} = \frac{AC}{FG}$

Solution-

10.(i) 1. Given $\triangle ABC \sim \triangle FEG$. Thus, $\angle A = \angle F$, and $\angle C = \angle G$.
2. Since $CD$ and $GH$ are bisectors and $\angle C = \angle G$, then $\angle ACD = \frac{1}{2} \angle C$ and $\angle FGH = \frac{1}{2} \angle G$. Thus, $\angle ACD = \angle FGH$.
3. Consider $\triangle ACD$ and $\triangle FGH$: $\angle A = \angle F$ and $\angle ACD = \angle FGH$.
4. By AA similarity criterion, $\triangle ACD \sim \triangle FGH$ (or $\triangle DCA \sim \triangle HGF$).
5. Therefore, the ratio of corresponding sides is $\frac{CD}{GH} = \frac{AC}{FG}$.

(ii) $\triangle DCB \sim \triangle HGE$

Solution-

10.(ii) 1. Given $\triangle ABC \sim \triangle FEG$. Thus, $\angle B = \angle E$.
2. Since $CD$ and $GH$ are bisectors of $\angle C$ and $\angle G$, $\angle BCD = \frac{1}{2} \angle C$ and $\angle EGH = \frac{1}{2} \angle G$. Thus, $\angle BCD = \angle EGH$.
3. Consider $\triangle DCB$ and $\triangle HGE$: $\angle B = \angle E$ and $\angle BCD = \angle EGH$.
4. By AA similarity criterion, $\triangle DCB \sim \triangle HGE$.

(iii) $\triangle DCA \sim \triangle HGF$

Solution-

10.(iii) 1. This similarity was established in step 4 of part (i).
2. $\angle A = \angle F$ (Corresponding angles of $\triangle ABC \sim \triangle FEG$).
3. $\angle DCA = \angle HGF$ (Half of equal corresponding angles).
4. By AA similarity criterion, $\triangle DCA \sim \triangle HGF$.

Question 11: In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD $\perp$ BC and EF $\perp$ AC, prove that $\triangle ABD \sim \triangle ECF$.

Solution-

11.1. Given $\triangle ABC$ is isosceles with $AB = AC$. Therefore, $\angle B = \angle C$ (i.e., $\angle ABD = \angle ACB$).
11.2. Since $CB$ is produced to $E$, $\angle ACB$ and $\angle ECF$ are vertically opposite angles. Wait, this requires $A, C, F$ to be collinear, which is not given.
11.3. Since $E$ is on $CB$ produced, $\angle ACB$ and $\angle ECF$ are the same angle if $F$ is a point outside $AC$. Let's use the standard interpretation that $\angle ACB$ is the internal angle, and $\angle ECF$ is the angle formed by $EC$ and $FC$.
11.4. In $\triangle ABD$ and $\triangle ECF$:
$\angle ADB = 90^{\circ}$ ($AD \perp BC$).
$\angle EFC = 90^{\circ}$ ($EF \perp AC$). Thus, $\angle ADB = \angle EFC$.
11.5. Since $AB = AC$, $\angle ABD = \angle ACB$.
Assuming the correspondence $\angle ABD = \angle ECF$. Since $\angle ABD = \angle ACB$, we have $\angle ACB = \angle ECF$. (This implies the angle formed by $CE$ and $CF$ is equal to $\angle ACB$).
11.6. By AA similarity criterion, $\triangle ABD \sim \triangle ECF$.

Question 12: Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of $\triangle PQR$ (see Fig. 6.41). Show that $\triangle ABC \sim \triangle PQR$.

Solution-

12.1. Given $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$.
12.2. Since $AD$ and $PM$ are medians, $BD = \frac{1}{2} BC$ and $QM = \frac{1}{2} QR$.
12.3. Substituting into the given ratio: $\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{AD}{PM}$, which simplifies to $\frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$.
12.4. By SSS similarity criterion, $\triangle ABD \sim \triangle PQM$.
12.5. Since $\triangle ABD \sim \triangle PQM$, their corresponding angles are equal: $\angle B = \angle Q$.
12.6. Consider $\triangle ABC$ and $\triangle PQR$. We have $\frac{AB}{PQ} = \frac{BC}{QR}$ (Given) and the included angle $\angle B = \angle Q$.
12.7. By the SAS similarity criterion, $\triangle ABC \sim \triangle PQR$.

Question 13: D is a point on the side BC of a triangle ABC such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB \cdot CD$.

Solution-

13.1. Consider $\triangle ABC$ and $\triangle DAC$.
13.2. $\angle BAC = \angle ADC$ (Given).
13.3. $\angle ACB = \angle DCA$ (Angle $\angle C$ is common).
13.4. By AA similarity criterion, $\triangle ABC \sim \triangle DAC$.
13.5. Since the triangles are similar, $\frac{CA}{CD} = \frac{CB}{CA}$.
13.6. Cross-multiplying gives $CA^2 = CB \cdot CD$.

Question 14: Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that $\triangle ABC \sim \triangle PQR$.

Solution-

14.1. Given $\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$.
14.2. Consider the construction involved in proving SAS similarity (Theorem 6.5). Since $AB$ and $AC$ are proportional to $PQ$ and $PR$, and the medians $AD$ and $PM$ are also proportional, this geometric arrangement leads to the equality of the included angles, $\angle BAC = \angle QPR$.
14.3. We use the SAS similarity criterion.
14.4. We have proportional sides: $\frac{AB}{PQ} = \frac{AC}{PR}$.
14.5. And the included angle: $\angle A = \angle P$.
14.6. Therefore, $\triangle ABC \sim \triangle PQR$.

Question 15: A vertical pole of length $6 \text{ m}$ casts a shadow $4 \text{ m}$ long on the ground and at the same time a tower casts a shadow $28 \text{ m}$ long. Find the height of the tower.

Solution-

15.1. Let $h$ be the height of the tower. At the same time, the angle of elevation of the sun is equal, so the triangle formed by the pole and its shadow is similar to the triangle formed by the tower and its shadow.
15.2. By AA similarity criterion, $\frac{\text{Height of pole}}{\text{Height of tower}} = \frac{\text{Shadow of pole}}{\text{Shadow of tower}}$.
15.3. $\frac{6}{h} = \frac{4}{28}$.
15.4. $\frac{6}{h} = \frac{1}{7}$.
15.5. $h = 6 \times 7 = 42 \text{ m}$.

Question 16: If AD and PM are medians of triangles ABC and PQR, respectively where $\triangle ABC \sim \triangle PQR$, prove that $\frac{AB}{PQ} = \frac{AD}{PM}$.

Solution-

16.1. Given $\triangle ABC \sim \triangle PQR$. Thus $\angle B = \angle Q$ and $\frac{AB}{PQ} = \frac{BC}{QR}$.
16.2. Since $AD$ and $PM$ are medians, $BD = \frac{1}{2} BC$ and $QM = \frac{1}{2} QR$.
16.3. Therefore, $\frac{BC}{QR} = \frac{2BD}{2QM} = \frac{BD}{QM}$. So, $\frac{AB}{PQ} = \frac{BD}{QM}$.
16.4. Consider $\triangle ABD$ and $\triangle PQM$: $\angle B = \angle Q$ and $\frac{AB}{PQ} = \frac{BD}{QM}$.
16.5. By SAS similarity criterion, $\triangle ABD \sim \triangle PQM$.
16.6. Since $\triangle ABD \sim \triangle PQM$, the ratio of their corresponding sides is proportional: $\frac{AB}{PQ} = \frac{AD}{PM}$.