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ABCD is a square and the diagonals  intersect at O. If P is a point on AB such that AO=AP , Prove that 3POB=AOP

Posted by: Abhishek in Geometry & Co-ordinate. 2 years ago, Total Answer(s): 1

Answer(s)

First that triangle AOP is an isosceles triangle (since AO=AP)
Therefore, angle AOP=45 degree
So, angle AOP= angle APO= (180-45)/2= 67.5 degree
Now,since angle APO+ angle BPO =180 degree
We have, angle BPO=180-67.5 = 112.5 degree
Next, angle PBO= 45 degree
We have,
angle POB= 180-(angle PBO)-(angle BPO)
       = 180-45-112.5=22.5 degree
So, 3* angle POB= 3*22.5 = 67.5 degree
     =angle AOP
Hence Proved

By: Dinesh, 2 years ago

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