Question

In the figure, AB and AC are two equal chords of a circle.then bisectors of $\angle BAC$ pass through the center of the circle.

• A. True
• B. False

Answer 1

The correct option is A True

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$\delta$APB and $\Delta$APC

AB=AC ...(Given)

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

AP=AP ....(common)

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

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

$\angle APB+\angle APC=180$

...(Linear pair)

$2\angle APB=180$

....$(\angle APB=\angle APC)$

$⟹\angle APB=90$

∘

∴AP is perpendicular bisector of chord BC.

Hence, AP passes through the center of the circle.