In the given figure, 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°–12∠A
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Ray BO is the bisector of angle CBE
Therefore,angle CBO=1/2of angle CBE
=1/2(180-y)
=90-y/2 (1)
Similarly,ray OC is the bisector of angle BCD
Therefore,angle BCO=1/2 of angle BCD
=1/2(180-z)
=90-z/2 (2)
In triangle BOC,angle BOC+BCO+CBO=180 (3)
Substituting (1,2,3) you get
Angle BOC+90-z/2+90-y/2=180
Angle BOC=z/2+y/2
=1/2(y+z)
But,x+y+z=180 (angle sum property)
y+z=180-x
Angle BOC=1/2(180-x)
=90-x/2
=90-1/2angle BAC
Please brainlest me the answer
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