bisectors of ∠B and ∠C of ∆ABC intersect at point P. Prove that ∠BPC = 90° + 1/2∠BAC.

Question

In the given figure, bisectors of ∠B and ∠C of ∆ABC intersect at point P. Prove that ∠BPC = 90° + \(\frac{1}2\)∠BAC.

Answer 1

In ∆ABC, 

∠BAC + ∠ABC + ∠ACB = 180° …[Sum of the measures of the angles of a triangle is 180°] 

∴ ∠BAC + – ∠ABC + ∠ACB = 180 … [Multiplying each term by \(\frac{1}2\) ]

∴ ∠BAC + ∠PBC + ∠PCB = 90° 

∴ ∠PBC + ∠PCB = 90° – 1 ∠BAC ………(i) 

In∆BPC, 

∠BPC + ∠PBC + ∠PCB = 180° …….[Sum of measures of angles of a triangle] 

∴ ∠BPC + 90° – \(\frac{1}2\)∠BAC = 180° ……[From (i)] 

∴ ∠BPC = 180° – 90°\(\frac{1}2\) ∠BAC 

= 180°- 90°+\(\frac{1}2\) ∠BAC 

= 90°+ \(\frac{1}2\)∠BAC