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Answer 1 · 2017-08-13T11:06:53

See below.

$[ABD]$ 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

$x = 20^@$

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NOTE:

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