Question

In the figure given below, O is the centre of the circle and PA and PB are tangent segments. Show that quadrilateral AOBP is cyclic.

Answer 1

Answer:

I am attaching the step wise solution of the given problem below. Check this solution and if you have any problem, drop a comment in the comment box.

As triangle are perpendicular to  the radius so

OA ⊥AP & OB ⊥BP

∠OAP = 90°, ∠OBP = 90°

∠OAP+  ∠OBP = 180°

In quadrilateral , OAPB

∠APB +∠AOB +∠AOP +∠OBP = 360

∠APB + ∠AOB + 180 = 360

∠APB + ∠AOB = 180

I hope this helps for you.

Answer 2

Solution:

A cyclic quadrilateral is one in which sum of opposite angles are supplementary.

ie. Sum of opposite angles are = 180°

As Point P is an external point and from P two tangents of equal length PA and PB are drawn at circle.

From the Theorem of tangent,that radius drawn an angle of 90°

Thus

< OBP = 90°

< OAP = 90°

< OBP + < OAP = 90°+ 90°


< OBP + < OAP = 180°

these are the opposite angles of cyclic quadrilateral,and sum of both are 180°, thus sum of remaining two angles are also 180°


< AOB + < BPA = 180°

hence OAPB is a cyclic quadrilateral.

Hope it helps you.