In a triangle ABC , the internal bisectors of angles B and C meet at P and the external bisectors of the angles B and C meet at Q. prove that:

angle BPC + angle BQC = 2 rt. angles.

AngleBPC+angleBQC=180 degrees

To prove that angle BPC + angle BQC = 2 right angles, we can start by drawing a diagram of triangle ABC.

First, let's find the measure of angle BPC. The internal bisectors of angles B and C meet at point P. This means that angle BPC is half the measure of angle B, since the internal bisector bisects angle B.

Similarly, let's find the measure of angle BQC. The external bisectors of angles B and C meet at point Q. This means that angle BQC is half the supplement of angle B, since the external bisector bisects the supplement of angle B. The supplement of angle B is the angle that, when added to angle B, equals 180 degrees. So, the supplement of angle B is 180 - angle B. Therefore, angle BQC is half of (180 - angle B).

Now, let's add angle BPC and angle BQC together:
angle BPC + angle BQC = (1/2) angle B + (1/2) (180 - angle B)
= (1/2) angle B + (90 - (1/2) angle B)
= (1/2) angle B + 90 - (1/2) angle B

Now, notice that the (1/2) angle B terms cancel out.
This simplifies the equation to:
angle BPC + angle BQC = 90 degrees

Therefore, angle BPC + angle BQC equals 90 degrees, which is a right angle.

To summarize, the internal bisector of angle B bisects angle B to form angle BPC, and the external bisector of angle B bisects the supplement of angle B to form angle BQC. By showing that angle BPC + angle BQC equals 90 degrees, we have proven that the sum of these two angles is equal to 2 right angles.