Question

the sides AB and AC of ∆ABC are produced to points E and D.If bisectors BO and CO of angle CBE and angle BCD meet at O,then prove that angle BOC=90°-1\2 angle BAC

Answer 1

Hi friend
Ray BO is the bisector of <CBE
Therefore, <CBO=1/2<CBE
=1/2(180°-y)
=90°-y/2 .......(1)
similarly ray CO is the bisector of <BCD
Therefore , <BCO=1/2<BCD
=1/2(180°-z)
=90°-z/2.......(2)

In ∆BOC,<BOC+<BCO+<CBO=180°.........(3)
substituting (1) & (2) in (3),you get
<BOC+90°-z/2+90°-y/2 =180°
so. <BOC=z/2+y/2
or. <BOC=1/2(z+y)............. (4)
But. x+y+z=180°
(Angel sum property of a triangle)
Therefore. y+z=180°-x
Therefore,(4) become
<BOC=12(180°-x)
=90°-x/2
=90°-1/2<BAC

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