in triangle ABC the internal bisector of angle B and angle C meet at the point P and the external bisectors of angle b and c meet at Q prove that angle BPC +BQC = 180 degree

∠ACB and ∠QCE form a linear pair

Hence ∠ACB+∠QCB =180º

⇒1/2∠ACB + 1/2∠QCB = 90°

⇒ ∠PCB+∠BCQ = 90°   ( As PC and QC are angle bisectors)

⇒ ∠PCQ = 90°

Similarly it can be proven that

∠PBQ = 90°

now in quadrilateral BPCQ, sum of all the four angles = 360º

⇒ ∠BPC+∠PCQ+∠CQB+∠QBP=360º

⇒ ∠BPC+∠BQC = 180º          ( as ∠PCQ=∠PBQ=90º)