In the given figure, O is the center of the circle. If chord AB = chord AC, OP ⊥ AB and OQ ⊥ AC; show that : PB = QC.

Let AB = AC = x

Given,

OP ⊥ AB and OQ ⊥ AC

⇒ OM ⊥ AB and ON ⊥ AC

Since, the perpendicular to a chord from the center 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 center,

∴ 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.