**ABCD is a cyclic quadrilateral. A line through D meets BC at E and AC at F. Then the circumcircles of triangles AFD and BDE are tangent to each other at D.**

**Proof**

I will just handle the case where E is on the opposite side of AD to C. You would technically have to consider the cases when E is between AD and C, B and C, and to the left of B, but they are almost identical proofs.

This is actually pretty trivial – I just thought it’s quite nice. ∠DBE = ∠DAF (same segment) – this is in fact enough to guarantee the circles are tangent to each other. To see this, choose G on BD extended such that FG is parallel to BE:

Then triangle DFG is obtained from triangle DEB by an enlargement of negative scale factor, centre D – their circumcircles are thus related in the same way and so they are tangential at D. Since ∠DBE = ∠DAF and ∠DBE = ∠DGF (alternate angles), ∠DAF = ∠DGF so A lies on the circumcircle of triangle DGF (converse of same segment) whence the result is established.

Alternatively, draw a line tangent to the circle DAF at D.

It is straightforward to show that this line is also tangent to circle BDE at D using the alternate segment theorem and its converse, from which the result also follows.

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