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.

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Solution

Given: AB and AC are chords of the circle with centre O. AB = AC, OP ⊥ AB and OQ ⊥ AC
To prove: PB = QC
Proof:
AB = AC (Given)
⇒ $\frac{1}{2} A B = \frac{1}{2} A C$
The perpendicular from the centre of a circle to a chord bisects the chord.
∴ MB = NC ...(i)
Also, OM = ON (Equal chords of a circle are equidistant from the centre)
and OP = OQ (Radii)
⇒ OP - OM = OQ - ON
∴ PM = QN ...(ii)
Now, in ΔMPB and ΔNQC, we have:
MB = NC [From (i)]
∠PMB = ∠QNC [90° each]
PM = QN [From (ii)]
i.e., ΔMPB ≅ ΔNQC (SAS criterion)
∴ PB = QC
(CPCT)

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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.

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.