Question

In the figure, AB and AC are two equal chords of a circle. Prove that the bisector of

the ∠BAC passes through the centre of the circle.​

Answer 1

Step-by-step explanation:

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

Prove: Center O lies on the bisector of the ∠BAC.

Construction: Join BC. Let the bisector on ∠BAC intersect BC in P.

Proof:

In△APB and △APC

AB=AC ...(Given)

∠BAP=∠CAP ...(Given)

AP=AP ....(common)

∴△APB≅△APC ...SAS test

⇒BP=CP and ∠APB=∠APC ...CPCT

∠APB+∠APC=180

∘

...(Linear pair)

∴2∠APB=180

∘

....(∠APB=∠APC)

⇒∠APB=90

∘

∴AP is perpendicular bisector of chord BC.

Hence, AP passes through the center of the circle.