In the above question 18 if internal bisector of Angle B and angle C intersect at P prove that angle P is equal to 90 degree and Angle BOC plus BPC Equal to 180⁰ pls tell m†*‡★★★★★

∠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]