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.
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)
$\Rightarrow \frac{1}{2} \mathrm{AB}=\frac{1}{2} \mathrm{AC}$
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)