Question

$AD$ is an altitude of an isosceles $△ABC$ in which $AB=AC$. show that
(i) $AD$ bisects $BC$
(ii) $AD$ bisects $\angle A$

Answer 1

Given:

In $\Delta ABC$, $AD\perp BC$.

To prove:

(i) $AD$ bisects $BC$.

(ii) $AD$ bisects $\angle A$

Proof:

In $\Delta ABD$ and $\Delta ACD$

$AB=AC........$(Given)
$AD=AD.......$(Common Side)

$\angle ADB=\angle ADC={90}^{\circ }$ ($\because AD\perp BC$)

By $RHS$ congruence rule

$\Delta ADB\cong \Delta ADC$

Since, the corresponding parts of the congruent triangles are equal.

$BD=DC$

$⟹AD$ bisects $BC$---(i) and

$\angle BAD=\angle CAD$

$⟹AD$ bisects $\angle A$ ---(ii)