In the following diagram, EB bisects AD. ACD is a isosceles triangle.

Prove that the angle CBE is 135 °

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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)

`=180^o-(180^o-(90^o +n))`

`=90^o +n`

`=angle AXB` (opposite angle)

Now consider `DeltaABX, angle ABX=180^o-(m+(90^o +n))`

`=(90^o-(m+n))`

Therefore, `angle CBE=180^o-angle ABX`

`=180^o-(90^o-(m+n))`

`=90^o +(m+n)`

Again, from `Delta CBE, angle CBE=180^o-(m+n)`

Hence `90^o +(m+n)=180^o -(m+n)`

`rArr (m+n)=90^o/2=45^o`

So, `angle CBE=180^o-45^o=135^o`

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