Two chords AB and AC of a circle are equal. Prove that the center of the circle, lies on the bisector of the angle BAC.
Two chords AB and AC of a circle are equal. Prove that the center of the circle, lies on the bisector of the angle BAC.

Selina Solutions Concise Class 10 Maths Chapter 18 ex. 18(C) - 4

Given: 

    \[AB\text{ }and\text{ }AC\]

are two equal chords of

    \[C\text{ }\left( O,\text{ }r \right)\]

Required to prove: Centre,

    \[O\]

lies on the bisector of 

    \[\angle BAC\]

Construction: Join

    \[BC\]

Let the bisector of 

    \[\angle BAC\]

intersects

    \[BC\text{ }in\text{ }P\]

Proof:

In 

    \[\vartriangle APB\text{ }and~\vartriangle APC\]

    \[AB\text{ }=\text{ }AC\]

[Given]

    \[\angle BAP\text{ }=~\angle CAP\]

[Given]

    \[AP\text{ }=\text{ }AP\]

[Common]

Hence,

    \[\vartriangle APB\cong \vartriangle APC\text{ }by\text{ }SAA\]

congruence criterion

So, by CPCT we have

    \[BP\text{ }=\text{ }CP\text{ }and~\angle APB\text{ }=~\angle APC\]

And,

    \[\angle APB\text{ }+~\angle APC\text{ }=\text{ }{{180}^{0}}\]

[Linear pair]

    \[2\angle APB\text{ }=\text{ }{{180}^{0}}\]

    \[~[\angle APB\text{ }=~\angle APC]\]

    \[\angle APB\text{ }=\text{ }{{90}^{0}}\]

Now,

    \[BP\text{ }=\text{ }CP\text{ }and~\angle APB\text{ }=\text{ }{{90}^{0}}\]

Therefore,

    \[AP\]

is the perpendicular bisector of chord

    \[BC\]

Hence,

    \[AP\]

passes through the centre,

    \[O\]

of the circle.