Because the circle intersects with AB at point D, CD and BC are both diameters of the circle, so the CDB angle is a right angle. As you can see, angle BDE=BCD, angle FDA is diagonal, and triangle ABC is isosceles, so angle CAD is equal to CBA and DF is perpendicular to AC, so FDA+CAD is 90, so we have said that CDB is right-angled, so in isosceles triangle.

BE:DE = DE； CE=BD:CD BD=6 CD=8, so we can find CE.

As for CF, you must have figured it out, so I won't talk about it. Then the angle e of SIN is CF:CE.

Right?