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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^@#

enter image source here

NOTE:

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