Question

AB and AC are two chords of a circle and the bisector of ∠BAC is a diameter of the circle. Prove that AB = AC.

Answer 1

Given: AB and AC are two chords of a circle and diametre of the circle is the bisector of ∠BAC.

To prove: AB = AC

Construction: Draw OF ⊥ AB and OE ⊥ AC.

Proof:

In ΔAFO and ΔAEO:

∠FAO = ∠EAO (As AD is the bisector of ∠BAC)

∠AFO = ∠AEO = 90° (By construction)

AO = AO (Common)

∴ ΔAFO ≅ ΔAEO (By Right-angle, hypotenuse and side criterion)

⇒ AF = AE (By c.p.c.t.)

We know that perpendicular from the centre of the circle to the chord bisects the chord.

∴ AF = FB and AE = EC

Now, AB = AF + FB

= 2AF

= 2AE

= AC (AE = )

Thus, chords AB and AC are equal.