Question

Please solve it....................

Answer 1

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.

Answer 2

Given :-

BP and CP are internal bisector of angle(B) and angle(C)

BQ and CQ are external bisector of angle(B) and angle(C)

_________________

We know that :-

If the bisector of Angles angle(ABC) and angle(ACB) meet a point O

then

=> angle(BOC) = 90 + 1/2[angle(A)]


In ΔABC

=> angle(BPC) = 90 + 1/2[angle(A)] __eq(1)

_______________

By using the theorem :-

sides AB and AC of a ΔABC are produced, and the external bisectors of angle(A) and angle(B) meet at point O

then.

=> angle(BOC) = 90 - 1/2[angle(A)]



Now, In ΔABC

=> angle(BQC) = 90 - 1/2[angle(A)]__eq(2)

___________

Adding eq(1) and eq(2)

angle(BPC) + angle(BQC) = 180

________[PROVED]