In traingle ABC, angle C = 40°. BQ and AR are the angle bisectors of angle B and angle A, respectively. Find traingle APQ.
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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
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