Question

From a point P outside a circle, with centre O, tangents PA and PB are drawn. Then, select the statements that are true.

• A. $\angle AOP=\angle BOP$
• B. OP is the $\perp$ bisector of chord AB.
• C. $\angle AOP=2\angle BOP$
• D. $\angle AOP=7\angle BOP$

Answer 1

Answer: (A), (B)

The correct options are
A

$\angle AOP=\angle BOP$

B

OP is the $\perp$ bisector of chord AB.

Given - A circle with centre O. A point P out side the circle. From P. PA and PB are the tangents to the circle. OP and AB are joined.

In $\Delta AOP$ and $\Delta BOP$,

AP = BP (Tangents from P to the circle)

OP = OP (Common)

OA= OB (Radii of the same circle)

$\therefore \Delta AOP\cong \Delta BOP$ (ssss postulate)

$\therefore \angle AOP=\angle BOP$ (C.P.C.T.)

Now in $\Delta OAM$ and $\Delta OBM$,

OA = OB (Radii of the same circle)

OM = OM (Common)

$\angle AOM=\angle BOM$ (Proved $\angle AOP=\angle BOP$)

$\therefore \Delta OAM\cong \Delta OBM$ (S.A.S. Postulate)

$\therefore$ AM = MB (C.P.C.T.)

and $\angle OMA=\angle OMB$ (C.P.C.T.)

But $\angle OMA+\angle OMB={180}^{\circ }$ (Linear pair)

$\therefore \angle OMA=\angle OMB={90}^{\circ }$

Hence OM or OP is the perpendicular bisector of AB.