In a ABC?, the sides AB and AC are produced to P and Q respectively. The bisectors of ? PBC and ? QCB intersect at a point O. Prove that ? BOC =90 ° ? 1/2 ?A

Answers (1)

Solution:  Let \angle PBC=2x   , \angle QCB=2y

ecause  OB  ,OC bisects the \angle PBC and  \angle QCB respectively 

	herefore angle OBP=angle OBC=2X/2=X\ \ Rightarrow angle OCB=OCQ=2y/2=y \ \

ecause in igtriangleup BOC,\ \ Rightarrow x+y+angle BOC=180 \ \Rightarrow x+y=180-angle BOC............(1)

\ecause angle ABC+angle PBC=180\ \ angle ABC=180-2x\ \

Similarly , \angle ABC=180-2y\ \

ecause In ecause igtriangleup ABC angle BAC+angle ABC+angle ACB=180^circ

\ x+y=90+(1/2)angle A

Now, \180-angle BOC=90+(1/2)angle A\ \Rightarrow angle BOC=90-(1/2)angle A

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