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.

[3 marks]
[RS Agrawal]
[Equal Chords of a Circle are Equidistant from the Centre]

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Solution

Solution:

AB = AC (Given)
⇒ ½ AB = ½ AC
The perpendicular from the centre of a circle to a chord
bisects the chord.
∴ MB = NC ...(i)
[1 mark]

Also, OM = ON (Equal chords of a circle are equidistant
from the centre) and OP = OQ (Radii)
⇒ OP − OM = OQ − ON
∴ PM = QN ...(ii)
[1 mark]

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)
[1 mark]

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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. [3 marks] [RS Agrawal] [Equal Chords of a Circle are Equidistant from the Centre]

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.

[3 marks]
[RS Agrawal]
[Equal Chords of a Circle are Equidistant from the Centre]