Question

AB and AC are two equal chords of a circle. Prove that the bisector of the angle BAC passes through the centre of the circle.

Answer 1

Given: We have a circle whose centre is O and chords AB and AC are equal. AM is the bisector of ∠BAC.

To prove: Centre O lies on the bisector of ∠BAC.

Construction: Join BM and CM.

Proof: In ΔBAM and ΔCAM,

AB = AC  ...[Given]

∠BAM = ∠CAM  ...[Given]

AM = AM  ...[Common]

∴ ΔBAM ≅ ΔCAM   ...[By SAS congruency]

`\implies` BM = CM  [By C.P.C.T.]  ...(i)

And ∠BMA = ∠CMA   [By C.P.C.T.]  ...(ii)

In ΔBOM and ΔCOM,

BM = CM   ...[By (i)]

OM = OM   ...[Common]

∠BMO = ∠CMO    ...[By (ii)]

∴ ΔBOM and ΔCOM   ...[By SAS congruency]

`\implies` ∠BOM = ∠COM  [By C.P.C.T.]  ...(iii)

Since, ∠BOM + ∠COM = 180°   ...(iv)

∴ By (iii) and (iv), ∠BOM = ∠COM = 90°

So, AM is the perpendicular bisector of the chord BC.

Thus, bisector of ∠BAC i.e., AM passes through the centre O.