Question

Question 4
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 AB and AC are two equal chords whose centre is O.
To prove that the centre O lies on the bisector of $\angle BAC$.
Construction : Join BC, draw bisector AD of $\angle BAC$

Proof
In $\Delta BAOand\Delta CAO$
AB = AC
$\angle BAO=\angle CAO$
AO = AO [common side]
$\therefore \Delta BAO\cong \Delta CAO$ [by SAS congruence rule]
$\Rightarrow$ BO = CO [by CPCT]
and $\angle BOA=\angle COA$ [by CPCT]
So, BO = CO, and $\angle BOA=\angle COA={90}^{\circ }$

So AD is the perpendicular bisector of the chord BC.
Hence the bisector of $\angle BAC$ i.e., AD passes through the centre O.