Question

ABC is a triangle inscribed in a circle. The bisectors of angle BAC , angle ABC and angle ACB meets the circle at X , Y and Z. Prove that in Triangle XYZ, angle YXZ =90° - angle BAC/2​

Answer 1

Step-by-step explanation:

Given△ABCisinscribedinC(0,r).

Thebisectorsof∠BAC,∠ABCand∠ACBmeetsthecircumcircle

of△ABC,inP,Q,Rrespectively.

InthefigureJoinRQ,

∠ABQ=∠APQ−(i)

{Anglesinthesamesegmentofacircleareequal}.

∠ABQ=∠QBC{BQisthebisectorof∠ABC}.

∴∠QBC=∠APQ−(ii)

Adding(i)&(ii)

∠ABQ+∠QBC=∠APQ+∠APQ

∴∠ABC=2∠APQ−(iii)

Similarily,∠ACB=2∠APR−(iv)

Adding(iii)&(iv)

∠ABC+∠ACB=2(∠APQ+∠APR)

∴∠ABC+∠ACB=2∠QPR−(v)

In△ABC,

∠ABC+∠BAC+∠ACB=180∘{Anglesumproperty}