In a triangle ABC, BD perpAC and CE perpAB. If BD and CE intersenct at O, prove that angle BOC = 180-A

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SORRY THE QUESTION IS :

 In a triangle ABC, BD is perpendicular to  AC and CE  is perpendicular to  AB. If BD and CE intersenct at O, prove that angle BOC = 180-A

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aec is a right angled triangle,

so angle c=90-a

name the intersection of bd and ac as m,

then, omc is a rt triangle

so angle boc=90-c

                       =90-(90-a)  [since c=90-a]

                      =180-a

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 Do u have maths exam tomorrow?? Just one advice... Just brush up ur knowledge about Trapeziums! and parallelograms! will help! :)

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