# 2015_Question_2

Claim: \triangle CDF is isoceles with CD=DF

Solution:
Proof by angle chasing (I assume there’s a way to prove this with lengths - perhaps power of a point?)

Set \angle FDC = x.
Then extend OD until it intersects with the larger circle again at a point H \neq D.

Since DF is a tangent, we know that \angle EDH = 90 - x
And by the inscribed angle theorem, this means that arc HE = 180-2x.

Subtracting this angle from arc HD = 180 we get that arc ED = 2x.
(This is the Tangent-Angle Theorem)

Now we can see that \angle DAE intersects arc DE with respect to the larger circle, meaning that it has an angle measure of x.

Since ABCD is a cyclic quadrilateral, opposite sides add up to 180 and thus \angle BCD = 180-x.

Finally, because supplementary angles add up to 180, \angle FCD = x and we are done.

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