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See diagram.

This problem can be solved by two methods.  One is by using fact that the angles subtended by a chord (or part of the circle) at any point on the circumference of the circle is same.  Second is by using the angle subtended at the center by a chord is twice the angle subtended at a point on the circumference.

Here let us the see the solution by the 1st method.

Mark angles BAD = A/2 = angle DAC;   angle ACF = angle FCB;
angle ABE = CBE = B/2

Chord FA subtends C/2 at C  => FDA = C/2
Chord AE subtends B/2 at B  => ADE = B/2

angle FDE = (B+C)/2 = (180-A)/2 = 90 - A/2

Chord BF subtends C/2 at C, so subtends same C/2 at E.
Chord BD subtends A/2 at A, so subtends A/2 at E.

angle FED =( A+C)/2 = (180-B)/2 = 90 -B/2

similarly the other one.

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