Question

In figure, arc AB is congruent to arc AC and O is the centre of the circle. Prove that OA is the perpendicular bisector of BC. [4 MARKS]

Answer 1

Concept: 2 Marks
Application: 2 Marks

Given: Arc AB is congruent to arc AC and O is the centre of the circle.

To prove: OA is the perpendicular bisector of BC.

Construction: Join OB and OC.

Proof:

Arc AB is congruent to arc AC [Given]

$\therefore$ chord AB = chord AC [If two arcs of a circle are congruent then their chords are also equal ]

$\therefore \angle AOB=\angle AOC$ …(i) [Equal chords of a circle subtend equal angles at the centre]

In $\Delta OBD$ and $\Delta OCD$,

$\angle DOB=\angle DOC$ [From (i)]

$OB=OC$ [Radii of the same circle]

$OD=OD$ [Common]

$\therefore \Delta OBD\cong \Delta OCD$ [By SAS]

$\therefore \angle ODB=\angle ODC$ [By C.P.C.T.]

$BD=CD$ [By C.P.C.T.]

But $\angle BDC={180}^{\circ }$

$\therefore \angle ODB+\angle ODC={180}^{\circ }$

$\Rightarrow 2\angle ODB={180}^{\circ }$

$\Rightarrow \angle ODB={90}^{\circ }$

$\therefore \angle ODB=\angle ODC={90}^{\circ }$

$\Rightarrow$ OA is the perpendicular bisector of BC.

Hence proved.