In circle O PA and PB are tangents The figure is not drawn t

In circle O, PA and PB are tangents, (The figure is not drawn to scale.) Prove that Delta APO = Delta BPO. Find m

Solution

a.

Circle O means the center of circle is O.

So, AO and BO are radius and are this equal.

PO is common in the two triangles APO and PBO

Also, AP and PB are the tangent to the circle. P is on extension of OD, which means that P is exactly at the center, if it was projected on a line segment AB. So, AP and PB tangent are equal

b.

AOP = POB ( as the 2 trinagles are congurent the 2 conguent angles are also equal)

So, POB = 64

Also, POB and BOD make 180, given they are on the same line segment PD

So, POB +BOD=180

64+BOD = 180

BOD = 180-64 = 116