Question

10. In the given figure, the internal bisector of B and external bisector of C meet at D. Prove that: D= 1/2 angle A​

Answer 1

$\huge\mathfrak\pink{answer}$

∠ABC+ext.∠∠ABC=180o (Angles on a straight line)

21(∠ABC+ext.∠ABC)=90o

∠PBC+∠QBC=90o (PB bisect Interior ∠B, QB bisects ext.∠B)

∠PBQ=90o

Similarly, ∠PCQ=90o

Sum of angles of quadrilateral PBCQ =360o

∠BPC+∠PBQ+∠PCQ+∠BQC=360o

∠BPC+∠BQC=180o

∴∠BPQ+∠BQC = 2 rt. angles