In the given figure, the perpendicular bisector of a chord . . .

Question : In the given figure, the perpendicular bisector of a chord AB of circle PXAQBY intersects the circle at P and Q. Prove that arc PXA ≅ arc PYB.

Doubt by Bhumika

Solution : 

Given : PQ is a perpendicular bisector of AB i.e. AM=BM & ∠PMA=∠PMB=90°

To prove : arc PXA ≅ arc PYB

Proof : 

In ΔPMA and ΔPMB

PM=PM (Common)
∠PMA=∠PMB (Each 90°)

AM=BM (Given)

ΔPMA ≅ ΔPMB (By SAS)

PA=PB (By CPCT)

arc PXA ≅ arc PYB [If two chords are equal, they subtend equal arcs]

Hence Proved