Question

S is the circumcircle of △ABC. Bisectors of ∠A, ∠B, ∠C meet S

at P, Q, R respectively. I is the incenter of △ABC.

1: Prove that I is orthocenter of △P QR.

2: Given m∠P = 70, m∠Q = 54, find m∠A.

3: Given BI = 6, BR = 8, BP = 10, find P R.

4: Given m∠P IR = 110, find m∠ABC.

Answer 1

Answer:

Solution

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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