Question

Two chords AB and AC of a circle are equal. Then, the centre of the circle, lies on the bisector of angle BAC.

• A. True
• B. False

Answer 1

The correct option is A

True

Given - A circle in which two chords AC and AB are equal in length. AL is the bisector of $\angle$ BAC.

In $\Delta$ ADC and $\Delta$ ADB,

AD=AD (Common)

AB=AC (Given)

$\angle$ BAD = $\angle$ CAD (Given)

$\therefore \Delta ADC\cong \Delta ADB$ (SAS postulate)

$\therefore$ BD = DC (C.P.C.T.)

and $\angle ADB=\angle ADC$ (C.P.C.T.)

But $\angle ADB+\angle ADC={180}^0$ (Linear pair)

$\therefore \angle ADB=\angle ADC={90}^0$

$\therefore$ AD is perpendicular bisector of chord BC.

$\because$ The perpendicular bisector of a chord passes through the centre of the circle.

$\therefore$ AD is the bisector of $\angle$ BAC passes through the centre O of the circle.