Question

AB and AC are equal chord of a circle . BC is produced to P so that CP = CA. If PA cuts the circle at Q, prove that BQ bisects angle ABC

Answer 1

Answer:

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.

Step-by-step explanation:

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