Question

In the adjoining figure, O is the centre of a circle. If AB and AC are chords of the circle such that AB = AC, OP ⊥ AB and OQ ⊥ AC, prove that PB = QC.

Answer 1

It is given that AB = AC

Dividing the equation by 2

We get

½ AB = ½ AC

Perpendicular from the centre of a circle to a chord bisects the chord

MB = NC ……. (1)

We know that the equal chords are equidistant from the centre of the circle

OM = ON and OP = OQ

Subtracting both the equations

OP – OM = OQ – ON

So we get

PM = QN ……… (2)

Consider △ MPB and △ NQC

We know that

∠PMB = ∠QNC = 90^o

By SAS congruence criterion

△ MPB ≅ △ NQC

PB = QC (c. p. c. t)

Therefore, it is proved that PB = QC.