Question

In a A ABC.
The sides AB and AC are produced to P
and Q respectively. If the bisectors of
PBC and ZQCB intersect at Apoint O.
Prove that ZBOC = 90°- ZA​

Answer 1

Step-by-step explanation:

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Answer 2

Refers to the attachments

GIVEN: /\ ABC in which side AB and AC are produced to P and Q respectively. The bisector of <PBC and <QCB intersected at O.

TO PROVE: <BOC=90°-1/2<A

PROOF: <ABC+ <CBP=180°(Linear pair)

=> <B + 2<1 =180°

=> 2<1 =180°- <B

=> <1=90°-1/2<B.......(i)

Again

<ACB+ <BCQ =180°

=> <C +2<2=180°

=> 2<2=180°- <C

=> <2 = 90° - 1/2<C........(ii)

In /\ BOC

<1 + <2 + <BOC =180°

90°-1/2(<B + <C ) + <BOC =180°[from i and ii]

<BOC =1/2(<B+ <C)

since, <A + <B + <C =180°

=> <B + <C =180°- <A

<BOC=1/2(180°- <A)

<BOC=90°-1/2<A