Question

In $△ABC$, $AB=AC$ and bisector of angle $A$ meets $BC$ at $D$, then prove that $AD$ is perpendicular to $BC$.

Answer 1

In $△ABD$ and $△ACD$,

$AD=AD$ [Common]
$\angle BAD=\angle CAD$ [$AD$ is the bisector of angle $A$]
$AB=AC$ [Given]

So, by $SAS$ rule of congruence,
$△ABD\cong △ACD$

As we know that, corresponding angles of congruent triangles are equal so,

$\angle ADC=\angle ADB=x$

Now,

$\angle ADC+\angle ADB={180}^{\circ }$ [Linear Pair]
$\Rightarrow 2x={180}^{\circ }$
$\Rightarrow x={90}^{\circ }$
$\Rightarrow \angle ADC=\angle ADB={90}^{\circ }$

Thus, $AD\perp BC$