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, $$\angle CBD = \angle EBO = \dfrac{1}{2} \angle CBE$$similarlyCO 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 $$\triangle OBC, \angle BOC + \angle BCO + \angle CBO = 180^{\circ}$$$$\angle BOC + \dfrac{1}{2}(x+y)+\dfrac{1}{2} (x+z) = 180^{\circ}$$$$\angle BOC + \dfrac{1}{2} (x+y+x+z) = 180^{\circ}$$$$x+y+z = 180^{\circ}$$$$\angle BOC + \dfrac{x}{2} + \dfrac{1}{2} \times 180^{\circ} = 180^{\circ}$$$$\angle BOC = 90^{\circ} - x/2$$Mathematics

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