Question

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

Answer 1

Prove the given condition :

Given,

1. $PXAQBY$ is the circle.
2. Perpendicular bisector of chord $AB$ intersects the circle at $P$ and $Q$.

$PQ$ is the perpendicular bisector of $AB$,

$\therefore AM=BM$

In $△APM$ and $△BPM$,

1. $AM=BM$ $[\because PM$ bisects $AB$ at $M]$
2. $\angle AMP=\angle BMP=90°$ $\left[\because PQ\perp AB\right]$
3. $PM$ is the common side

By SAS congruence rule, if any two sides and anangle included between the sides of one triangle are equivalent to the corresponding two sides and the included angle between the sides of the second triangle, then the two triangles are said to be congruent.

$\therefore △APM\cong △BPM$

Since, corresponding parts of congruent triangles are congruent, we can say that $AP=BP$

If two chords of a circle are equal, then their corresponding arcs are congruent i.e.$\mathrm{arc}PXA\cong \mathrm{arc}PYB$

Hence proved that $\mathrm{arc}PXA\cong \mathrm{arc}PYB$.