Question

Q
40. A triangle ABC is inscribed in a circle. The
bisectors of angles BAC, ABC and ACB meet
the circumcircle of the
А.
triangle at points P, Q and
R
R respectively. Prove that :
(i) ZABC = 2 ZAPQ,
(ii) ZACB = 2ZAPR,
B
С
To
1
od
) =
2
olup
Ż ZBACO​

Answer 1

• Answer:

ccv

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Answer 2

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