The side AC of triangle ABC, IS produced to E. The bisector of angel B meets AC in D, then angleBAC+angleBCE=
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In △ABC (Fig.6.32), the sides AB and AC of △ABC are produced to points E and D respectively.If bisectors BO and CO of ∠CBE and ∠BCD respectively meet at point O.then prove that ∠BOC = 90 degree - 1/2∠A.
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