In the figure (ii) given below, O is the center of a circle. If AB and AC are chords of the circle such that AB = AC and OP ⊥ AB, OQ ⊥ AC, prove that PB = QC.

Let AB = AC = x

Given,

OM ⊥ AB and ON ⊥ AC

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

∴ AM = MB = $\dfrac{x}{2}$

and

AN = NC = $\dfrac{x}{2}$

∴ MB = NC ……….(1)

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

∴ ON = OM = y (let).

Let radius of circle be r.

From figure,

OQ = OP = r

QN = OQ - ON = r - y

PM = OP - OM = r - y

∴ QN = PM ……….(2)

In △QNC and △PMB,

NC = MB [From (1)]

QN = PM [From (2)]

∠QNC = ∠PMB (Both equal to 90°)

△QNC ≅ △PMB by SAS axiom.

∴ PB = QC (By C.P.C.T.)

Hence, proved that PB = QC.