In traangle ABC , AB =AC and E is any point on the extended BC . If the circumcircle of `DeltaABC` intersect AE at the point D then that `angleACD=angleAEC`.

Explanation

In `DeltaABC`, AB=AC,
`thereforeangleABC=angleACB` ………….(1)
Now , ABCD is a cyclic quadrilateral,
`thereforeangleABC+angleADC=180^(@)`
or , `angleACB+angleADC=180^(@)` ……..(2) [from (1)] ,
Again , `angleACB+angleACE=1" straight angles " =180^(@)` ........(3)
From (2) and (3) we get , `angleACB+angleADC=angleACB+angleACE`
or, `angleADC=angleACE" or " angleDCE+angleDEC=angleACD+angleDCE" or " angleDEC=angleACD`
`thereforeangleACD=angleDEC`
Hence `angleACD=angleAEC`