Question

AB and AC are equal chords of a circle with centre O. Prove that OA bisects ∠BAC.

Answer 1

AP is the perpendicular bisector of the chord BC and AP

Explanation:

Let the two chords AB and AC are chords.

Centre of the circle O lies on the bisector of the ∠BAC.

Join BC

Let the bisector of ∠BAC intersect BC at P.

From ΔAPB and ΔAPC, we get:

AB =AC

BAP =CAP

AP=AP

ΔAPB congruent to ΔAPC

So BP = CP and ∠APB =∠APC.

Now ∠APB + ∠APC = 180° as they are a linear pair.

2*∠APB = 180°

∠APB = 180° / 2

∠APB = 90°

So BP = CP

Therefore AP is the perpendicular bisector of the chord BC and AP and passes through the centre of the circle O. Hence proved.

Answer 2

Answer:

Let the two chords AB and AC are chords.

Centre of the circle O lies on the bisector of the ∠BAC.

Join BC

Let the bisector of ∠BAC intersect BC at P.

From ΔAPB and ΔAPC, we get:

AB =AC

BAP =CAP

AP=AP

ΔAPB congruent to ΔAPC

So BP = CP and ∠APB =∠APC.

Now ∠APB + ∠APC = 180° as they are a linear pair.

2*∠APB = 180°

∠APB = 180° / 2

∠APB = 90°

So BP = CP

Therefore AP is the perpendicular bisector of the chord BC and AP and passes through the centre of the circle O. Hence proved.