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1.In a triangle ABC, the internal bisectors of angles Band C meet at P and the external bisectors of the angles Band C meet at Q.

Prove that : angle BPC+ angle BQC=2 rt. angles




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See the diagram.
Concentrate on the colored areas  and angles.

ABC is the triangle.  BE and CF are the internal bisectors of angles B and C respectively.  They intersect at P.

Angles ABG and ACH are the exterior angles at B and C to the triangle ABC.  Now,
draw the bisectors to those angles as shown.  BQ and CQ are the bisectors.

Join  PQ.  now look at the two triangles,  PBQ and PCQ.

angle CBG  = 180 = 2 * angle B/2 +  2 * angle y = 2 * (angle B/2 + angle y)
                = 2 * angle  PBH
      => angle PBH = 90 deg.

Similarly, angle BCD = 180 deg = angle BCA + angle ACD
                   = 2 * angle C/2 + 2 * angle ACK    =2 * ( C/2 + x)
                   = 2 * angle PCK 
        =>  angle PCK = 90 deg.

   angle PBH = exterior angle to triangle PBG = angle BPQ + angle BQP
                     = 90 deg.
  angle PCK = exterior angle to triangle PCQ = angle CPQ + angle CQP
             = 90 deg.

   now add angle PBH + angle PCK = 180 deg
               = angle BPQ + angle CPQ + angle BQP + angle CQP
                   = angle BPC + angle BQC

hence proved.
1 5 1
I hope it is helpful and you understand it now 
2 5 2
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