Question

If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that ae PXA $\cong$ Arc PYB.

Answer 1

As $PQ$ is the perpendicular bisect of $AB$.

Refer image,

So, $AM=BM$

In $\Delta APM$ and $\Delta BPM$, we have

$AM=BM$ [Proved above]

$\angle AMP=\angle BMP$ [Each $={90}^0$]

$PM=PM$ [Common side]

$\therefore \Delta APM\cong \Delta BPM$ [By SAS congruence rule]

So, $AP=BP$ [By C.P.C.T]

Hence, arc $PXA=ArcPYB$

(If two chords of a circle are equal, then their corresponding arcs are congruent)