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Question

In Figure, the sides AB and AC of $$\Delta$$ABC are produced to points E and D respectively. If bisectors BO and CO of $$\angle$$CBE and $$\angle$$BCD respectively meet at point O, then prove that $$\angle$$BOC$$=90^o=\dfrac{1}{2}\angle$$BAC.


Solution

REF. Imge.
BD is the bisector of $$ \angle CBE $$
So, $$ \angle CBD = \angle EBO = \dfrac{1}{2} \angle CBE $$
similarly
CO is the bisector of $$ \angle BCO $$
So, $$ \angle BCO = \angle DCO = \dfrac{1}{2} \angle BCD $$
$$ \angle CBE$$ is the exterior angle of $$\triangle ABC $$
Hence $$ \angle CBE = x+z \Rightarrow  \dfrac{1}{2} \angle CBC = \dfrac{(x+z)}{2}$$
$$ \angle CBO = \dfrac{1}{2}(x+z)$$
$$ \angle BCD $$ is the exterior angle  of $$ \triangle ABC $$
$$ \angle BCD = x+y $$
$$ \dfrac{1}{2} \angle BCD = \dfrac{1}{2} (x+y)$$
$$ \angle BCO = \dfrac{1}{2} (x+y)$$
in $$ \triangle OBC, \angle BOC + \angle BCO + \angle CBO = 180^{\circ}$$
$$ \angle BOC + \dfrac{1}{2}(x+y)+\dfrac{1}{2} (x+z) = 180^{\circ}$$
$$ \angle BOC + \dfrac{1}{2} (x+y+x+z) = 180^{\circ}$$
$$ x+y+z = 180^{\circ}$$
$$ \angle BOC + \dfrac{x}{2} + \dfrac{1}{2} \times 180^{\circ} = 180^{\circ}$$
$$ \angle BOC = 90^{\circ} - x/2 $$

1205267_1069457_ans_9f400af366744caca2f5fea3fe16f7f3.jpg

Mathematics

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