Question

ABCD is a parallelogram, produce DC and AB to E and F respectively. CP and BP are two external bisector of angels cCand B respectively. Which interests at P. Prove that angle BPC= 90 degrees

Answer 1

To prove that angle BPC is equal to 90 degrees in parallelogram ABCD, we can use congruent angles and the fact that CP and BP are external bisectors.

Step-by-step explanation:

In parallelogram ABCD, we extend DC and AB to points E and F respectively. Let CP and BP be the external bisectors of angles C and B respectively, which intersect at point P. We want to prove that angle BPC is equal to 90 degrees.

From the given information, we can observe that the angles BCD and ACD are congruent since they are opposite angles of a parallelogram. Similarly, the angles BAC and DAC are congruent.

Now, since CP is an external bisector of angle C, it splits angle DCA into two equal angles. Similarly, BP splits angle DBA into two equal angles.

Since the angles DCA and DBA are congruent, and the bisectors CP and BP split them into two equal angles, we can conclude that angle BPC is equal to 90 degrees.