Question

Prove that the right bisector of a chord of a circle, bisects the corresponding arc of the circle.

Answer 1

Let AB be a chord of a circle having its centre at O.

Let PQ be the right bisector of the chord AB, intersecting AB at L and the circle at P and Q.

Since the right bisector of a chord always passes through the centre, so PQ must pass through the centre O.

Join OA and OB.

$OA=OB$ [Each equal to the radius]

$\angle ALO=\angle BLO$ [Each equal to ${90}^{\circ }$]

$OL=OL$ [Common]

$\therefore \Delta OAL\cong \Delta OBL$ [By RHS congruency criterion]

$\Rightarrow \angle AOL=\angle BOL$ [C.P.C.T]


$\Rightarrow \angle AOQ=\angle BOQ$

$\therefore AQ=BQ$ [Arcs subtending equal angles at the centre are equal]