Exercise 7.2: Triangles — NCERT Solutions

Given: $AD$ bisects $\angle A$ (so $\angle BAD = \angle CAD$) and $AD \perp BC$ (so $\angle ADB = \angle ADC = 90^\circ$).

To prove: $AB = AC$.

Proof:

In $\triangle ABD$ and $\triangle ACD$:

1. $\angle BAD = \angle CAD$ (given)
2. $AD = AD$ (common side)
3. $\angle ADB = \angle ADC = 90^\circ$ (given)

By the ASA congruence rule, $\triangle ABD \cong \triangle ACD.$

By CPCT, $AB = AC,$ so $\triangle ABC$ is isosceles.

Given: $\angle B = \angle C$ and $\angle BAD = \angle CAD$.

To prove: $BD = CD$.

Proof:

In $\triangle ABD$ and $\triangle ACD$:

1. $\angle B = \angle C$ (given)
2. $\angle BAD = \angle CAD$ (given)
3. $AD = AD$ (common side)

Here two angles and a non-included side are equal, so by the AAS congruence rule, $\triangle ABD \cong \triangle ACD.$

By CPCT, $BD = CD.$

Given: $l$ bisects $\angle A$, so $\angle PAB = \angle QAB$. Also $BP \perp AP$ and $BQ \perp AQ$, so $\angle APB = \angle AQB = 90^\circ$.

To prove: $BP = BQ$.

Proof:

In $\triangle APB$ and $\triangle AQB$:

1. $\angle APB = \angle AQB = 90^\circ$ (given)
2. $\angle PAB = \angle QAB$ (l bisects $\angle A$)
3. $AB = AB$ (common side)

By the AAS congruence rule, $\triangle APB \cong \triangle AQB.$

By CPCT, $BP = BQ.$

Hence $B$ is equidistant from both arms of the angle.