Question

in figure the sides AB and AC are produced to point E and D respectively. if bisectors BO and CO of angleCBE and angleBCD respectively meet at point O, then prove they angleBOC =90° -1/2 angleBAC.
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Answer 1

Ray BO is the bisector of angle CBE

angle CBO=angle EBO-1\2 angle CBE

angle CBO =1\2 (180°-y)

angle CBO=90°-y\2

Ray CO is bisector of angle BCD

angle BCO=angle OCD-1\2 angle BCD

angle BCO=1\2(180°-z)

angle BCO=90°-z\2

In ∆BCO

angle BCO + angle BOC + angle CBO=180°

90°-z\2 + angle BCO + 90°-y\2=180°

angle BCO=z\2+y\2

angle BCO=y+z\2———(eq.1)

In ∆ABC

angle A + angle B + angle C=180°

x + y + z=180°

y+z=180°-x———(eq.2)

angle BCO=180°-x\2

angle BCO=90°-x\2

angle BCO=90°-angle BAC\2

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