Question

Question 2
If the perpendicular bisector of a chord AB of a circle PXAQBY intersect the circle at P and Q, prove that arc PXA $\cong$ arc PYB.

Answer 1

Let AB be a chord of a circle having centre at O.PQ be the perpendicular bisector of the chord AB, which intersects AB at M and it always passes through O.

To prove that, arc PXA $\cong$ arc PYB
Construction : Join AO and BO

Proof
In $\Delta$ APM and $\Delta$BPM
AM = MB [PM bisects AB]
$\angle PMA=\angle PMB$
$\left[PMistheperpendicularbisectorofAB\right]$
PM = PM [common side]
$\Delta APM\cong \Delta BPM$ [by SAS congruence rule]
PA = PB [by CPCT]
Therefore, Arc PXA $\cong$ Arc PYB