Question

A triangle ABC is inscribed in a circle and the bisectors of the angles meet the circumference at X,Y and Z. Show that the angles of the triangle are respectively 90°-X/2 , 90°-Y/2 and 90°-Z/2.​

Answer 1

Answer:

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}

∠ABC+∠ACB=180

∘

−∠BAC−(vi)

From(v)&(vi),weget

180

∘

−∠BAC=2∠QPR

∴∠QPR=

2

1

(180−∠BAC)

∠QPR=

2

1

×180

∘

−

2

1

∠BAC

⟹∠QPR=90−

2

1

∠BAC

Step-by-step explanation:

hope it will help you...