Question

$AB$ and $AC$ are two equal chords of a circle. Prove that the bisector of the $\angle BAC$ passes through the centre of the circle.

Answer 1

Given: $AB$ and $CD$ are two equal chord of the circle.

Prove: Center $O$ lies on the bisector of the $\angle BAC$.

Construction: Join $BC$. Let the bisector on $\angle BAC$ intersect $BC$ in $P$.

Proof:

In$△APB$ and $△APC$

$AB=AC$ ...(Given)

$\angle BAP=\angle CAP$ ...(Given)

$AP=AP$ ....(common)

$\therefore △APB\cong △APC$ ...SAS test

$\Rightarrow BP=CP$ and $\angle APB=\angle APC$ ...CPCT

$\angle APB+\angle APC={180}^{\circ }$ ...(Linear pair)

$\therefore 2\angle APB={180}^{\circ }$ ....$(\angle APB=\angle APC)$

$\Rightarrow \angle APB={90}^{\circ }$

$\therefore AP$ is perpendicular bisector of chord $BC$.

Hence, AP passes through the center of the circle.