In `Delta ABC,AB= AC and E` is any point on the extended BC. The circumeircel of `Delta ABC` intersects AE at the point D. Prove that ` angle ACD= angle AEC`.

Explanation

Give `Delta aBC, AB=AC` and E is any point on the produced BC. The circumcricle of `Delta ABC` intersects AE at the point D.
To prove : We have to prove that `angle ACD= angle AEC`
Construction : Let us jin B,D
Proof : External `angle ACB of Delta ACE=` internally opposite
`( angle AEC+ angle CAE=angle AEC+ angle CBD)`
`[:.` the angles produed by `CD, angle CAD= angle CBD]`
or, `angle ACB-, angle CBD= angle AEC or, angle ABC- angle CBD= angle AEC[ :. AB= AC, :. angle ACB= angle ABC]`
`or, angle ABD= angle AEC` [ as per figure]
`or, angle ACD= angle AEC [ :. ` the angle in circle proeduced by `AD, angle ABD= angle ACD]`
`:. angle ACD= angle AEC`