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In the following diagram, EB bisects AD. ACD is a isosceles triangle.Prove that...
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Label the point of intersection of EB and AD as X.
`angle EXD=180^o-angle BXD`
`=180^o-angle AXE` (opposite angle)
`=angle AXB` (opposite angle)
Now consider `DeltaABX, angle ABX=180^o-(m+(90^o +n))`
Therefore, `angle CBE=180^o-angle ABX`
Again, from `Delta CBE, angle CBE=180^o-(m+n)`
Hence `90^o +(m+n)=180^o -(m+n)`
So, `angle CBE=180^o-45^o=135^o`
Posted by llltkl on August 21, 2013 at 7:04 AM (Answer #1)
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