In the given figure, AB and AC are equal chords of a circle with centre O and OP ⟂ AB, OQ ⟂ AC. Prove that PB = QC.

Let AB = AC = x.

Given,

OM ⊥ AC and OL ⊥ AB.

Since, the perpendicular to a chord from the centre of the circle bisects the chord.

∴ AM = MC = $\dfrac{x}{2}$

and

AL = LB = $\dfrac{x}{2}$

∴ MC = LB ……(1)

Since, equal chords of a circle are equidistant from the centre,

∴ OM = OL = y (let).

Let radius of circle be r.

From figure,

OQ = OP = r

QM = OQ - OM = r - y

PL = OP - OL = r - y

∴ QM = PL …..(2)

In △QMC and △PLB,

MC = LB [From (1)]

QM = PL [From (2)]

∠QMC = ∠PLB (Both equal to 90°)

△QMC ≅ △PLB by SAS axiom.

∴ PB = QC (By C.P.C.T.)

Hence, proved that PB = QC.