Exercise 7.1: Triangles — NCERT Solutions

Given: $AC = AD$ and $AB$ bisects $\angle A$, so $\angle CAB = \angle DAB$.

To prove: $\triangle ABC \cong \triangle ABD$.

Proof:

In $\triangle ABC$ and $\triangle ABD$:

1. $AC = AD$ (given)
2. $\angle CAB = \angle DAB$ (since $AB$ bisects $\angle A$)
3. $AB = AB$ (common side)

Therefore, by the SAS congruence rule, $\triangle ABC \cong \triangle ABD.$

Since corresponding parts of congruent triangles are equal (CPCT), $BC = BD.$

Given: $AB = PQ$, $BC = QR$, $CA = RP$.

Since all three pairs of corresponding sides are equal, by the SSS congruence rule, $\triangle ABC \cong \triangle PQR.$

By CPCT, the corresponding angles are equal: $\angle A = \angle P,\quad \angle B = \angle Q,\quad \angle C = \angle R.$

Note: the angle equal to $\angle A$ is the one opposite the side $BC$ (namely $\angle P$, opposite $QR$), so matching must follow the vertex order carefully.

Given: $AB$ and $CD$ bisect each other at $O$, so $AO = BO$ and $CO = DO$.

To prove: $AC = BD$ and $AC \parallel BD$.

Proof:

In $\triangle AOC$ and $\triangle BOD$:

1. $AO = BO$ (O bisects $AB$)
2. $\angle AOC = \angle BOD$ (vertically opposite angles)
3. $CO = DO$ (O bisects $CD$)

By the SAS congruence rule, $\triangle AOC \cong \triangle BOD.$

By CPCT, $AC = BD$.

Also by CPCT, $\angle OAC = \angle OBD$. These are alternate interior angles for lines $AC$ and $BD$ with transversal $AB$, so $AC \parallel BD.$