Question

Answer ???

Answer 1

See below.

Explanation:

`[ABD]` is an isosceles triangle
`[ABE]` is an isosceles triangle also

then calling

`hat(BAE) = a`
`hat(BEA) = a`
`hat(CAD)= b = 10^@`
`hat(BDA)=d`
`hat(BAD)=d`
`hat(AFE)=c`

we have for triangles `[AFE]` and `[BFD]` respectively

`{(b+c+a=180^@),(x+c+d = 180^@),(d=a-b):}`

now substituting and subtracting the corresponding sides we have

`x -2b=0` or

`x = 20^@`

NOTE:

This can be quickly concluded also by considering over the circle, that the angle `hat(CAD) = hat(EBC)/2`