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  • If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA congruent to Arc PYB.

If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA ≅ Arc PYB.

Answers (1)

Given that the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q

Let AB be the chord having a centre at O.

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

Join AP and BP

In \triangle APM and \triangle BPM

AM = MB (PM is the perpendicular bisector of AB)

PM = PM (Common)

\angle PMA = \angle PMB  (90° each)

\therefore \triangle APM \cong \triangle BPM (SAS congruence)

PA = PB  (by CPCT)

If 2 chords of a circle are equal, then corresponding arcs are congruent

Hence, arc PXA \cong  arc PYB.

Hence proved

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