Natasha Circle Proof

a)

<PAO = <PBO since tangents meet radii at right angles

AO = BO   both are radii

PO = PO

Then by HL, PA = PB

Then, by SSS  .......ΔAPO ≅ ΔBPO

b) <AOD = <BOD = 180 - <AOP =  180 - 64 = 116°   [<AOP + <AOD = 180. And <AOD = <BOD by symmetry. Thus (180 - <AOP) = (180 - <BOP) = <BOD = 116°]

A. The tricky part of this proof is that we must recognize that angles PAO and PBO are both right angles because they are formed by radii to tangents.  Therefore they are congruent.

Segment AO is congruent to segment BO as they are both radii.

Segment PO is congruent to itself by the reflexive property of congruence. This essentially means that it is a segment that is a side length of both triangles.  We should also note that is the hypotenuse. 

With these pieces of information hand we are able to say the triangles are congruent by the Hypotenuse Leg theorem of triangle congruence!

b.  Since we proved the triangles congruent in part A it follows that angle POB has a measure of 64.  POB is supplementary to angle BOD (they sum to 180 degrees) there fore the measure of BOD is 180-62=114 degrees

Thank You! :)