Question

In a triangle ABC the internal bisectors of angle B and angle C meet at P and the external bisectors of angle B and angle C meet at Q. Prove that angle BPC plus angle BQC equal to 180

Answer 1

Answer:

∠ACB  and ∠QCB form a linear pair,

So, ∠ACB + ∠QCB = 180º

=> ∠ACB/2 + ∠QCB/2 = 180º /2

=> ∠ACB/2 + ∠QCB/2 = 90º                    

Again since PC and QC are the angle bisectors

=> ∠PCB + ∠BCQ = 90º

=> ∠PCQ = 90º            

Similarlry,

∠PBQ = 90º

In qradrilateral BPCQ,

Sum of all four angles = 360º

=> ∠BPC + ∠PCQ + ∠CQB + ∠QBP = 360º

=> ∠BPC + ∠BQC + 180º = 360º

=> ∠BPC + ∠BQC = 360º - 180º

=> ∠BPC + ∠BQC = 180º

=> ∠P = ∠Q = 90º                 [Since ∠PCQ = ∠PBQ = 90º]

Answer 2

Answer:

Consider a triangle ABC, BP and CP are internal bisectors of $\angle B$ and $\angle C$ respectively.

We need to prove that $\angle BPC+\angle BQC=180^0.$

We know that if the bisectors of the angles $\angle ABC$,  $\angle ACB$ meet at a point 0 the $\angle BOC=90^0+\frac{1}{2} \angle A.$

From $\triangle ABC$, we have  

$\angle BPC=90^0+\frac{1}{2} \angle A$.------------------------( 1 )

Using the theorem, "If the sides AB and AC of a triangle ABC are produced, and the external bisectors of $\angle B$ and $\angle C$ meet at 0, then

$\angle BOC=90^0+\frac{1}{2} \angle A$."

Thus from  $\triangle ABC$, we get

$\angle BQC=90^0-\frac{1}{2} \angle A$------------------------( 2 )

Add equation ( 1 ) and equation ( 2 ), and we get

$\angle BPC+\angle BQC=90^0+\frac{1}{2} \angle A+90^0-\frac{1}{2} \angle A$,

$\angle BPC+\angle BQC=180^0$,

Therefore, $\angle BPC+\angle BQC=180^0$.

Hence proved.