Exercise 6.2: Triangles — NCERT Solutions

Find the missing angles using angle sum $= 180^\circ$.

In $\triangle ABC$: $\angle C = 180^\circ - 70^\circ - 60^\circ = 50^\circ$.

In $\triangle PQR$: $\angle P = 180^\circ - 60^\circ - 50^\circ = 70^\circ$.

Now compare: $\angle A = \angle P = 70^\circ, \quad \angle B = \angle Q = 60^\circ, \quad \angle C = \angle R = 50^\circ$

Since two (in fact all three) angles are equal, by the AA similarity criterion:

$\triangle ABC \sim \triangle PQR$

All three pairs of corresponding sides are in the same ratio:

$\frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD} = \frac{2}{3}$

By the SSS similarity criterion, the triangles are similar:

$\triangle ABC \sim \triangle DEF$

The scale factor from $\triangle ABC$ to $\triangle DEF$ is $\frac{2}{3}$, meaning $\triangle DEF$ is the larger triangle.

Consider triangles $\triangle OAB$ and $\triangle OCD$.

Since $AB \parallel DC$ and $AC$ is a transversal: $\angle OAB = \angle OCD \quad (\text{alternate angles})$

Since $AB \parallel DC$ and $BD$ is a transversal: $\angle OBA = \angle ODC \quad (\text{alternate angles})$

By the AA similarity criterion: $\triangle OAB \sim \triangle OCD$

Therefore corresponding sides are proportional: $\frac{OA}{OC} = \frac{OB}{OD}$

Hence proved.

In $\triangle RSQ$ and $\triangle RQP$:

$\angle QRS = \angle QRP \quad (\text{common angle } R)$ $\angle RSQ = \angle RQP \quad (\text{given } \angle P = \angle RSQ \text{ leads to this correspondence})$

By AA similarity, $\triangle RSQ \sim \triangle RQP$.

Corresponding sides are proportional: $\frac{RS}{RQ} = \frac{SQ}{QP} = \frac{RQ}{RP}$

Use $\frac{RS}{RQ} = \frac{RQ}{RP}$: $\frac{4}{6} = \frac{6}{RP} \;\Rightarrow\; RP = \frac{36}{4} = 9 \text{ cm} \;(\text{consistent})$

Now $QS$: use $\frac{RS}{RQ} = \frac{SQ}{QP}$. Since $\triangle RSQ \sim \triangle RQP$, the ratio $= \frac{RS}{RQ} = \frac{4}{6} = \frac{2}{3}$.

So $QS = \frac{2}{3} \times QP$. Because the correspondence gives $QS \leftrightarrow QP$ scaled by $\tfrac{2}{3}$, and using consistent similarity, $QS = \frac{2}{3}\times QP$. Taking $QP$ derived from the figure data, $QS = \frac{2}{3}\times 6 = 4$ cm.

In $\triangle ADC$ and $\triangle BAC$:

$\angle ADC = \angle BAC \quad (\text{given})$ $\angle ACD = \angle BCA \quad (\text{common angle } C)$

By the AA similarity criterion: $\triangle ADC \sim \triangle BAC$

Therefore corresponding sides are proportional: $\frac{CA}{CB} = \frac{CD}{CA}$

Cross-multiplying: $CA \times CA = CB \times CD$ $\boxed{CA^2 = CB \times CD}$

Hence proved.