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In parallelogram ABCD, DC is extended to point E, and AB is extended to point F. CP and BP are the external bisectors of angles C and B, respectively, and they intersect at point P. Prove that angle BPC is equal to 90 degrees.

Answer :

Final Answer:

The angle BPC in parallelogram ABCD, where CP and BP are external bisectors of angles C and B respectively, is proven to be equal to 90 degrees.

Explanation:

In parallelogram ABCD, extend DC to point E and AB to point F.

Opposite angles in a parallelogram are equal, so angle BCD equals angle ADC.

External bisectors CP and BP are drawn at angles C and B, intersecting at point P.

Due to the angle bisector property, angles BPC and BPD are equal, as are angles APC and APD.

Angles APC, BPC, BPD, and APD together form a linear pair, and their sum is equal to 180 degrees.

Since angle BPD is a right angle (as it is a bisected angle), the remaining angle, BPC, must also be 90 degrees.

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