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Cherchez le quadrilatère cyclique: What is this about?
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The applet is supposed to illustrate the following problem:
Let ABCD be a cyclic quadrilateral. At A and D construct equal angles BAE and CDF in directions opposite relative to BC. Let E and F lie on BC. Then quadrilateral AEFD is cyclic,  EAF = EDFCAF = BDE. |
In a slightly different form the problem appeared in [Andreescu & Gelca, p. 9, problem 1.2.7]:
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Points B and C are given on the side BC of a convex quadrilateral AEFD (with B closer to E than to F.) It is known that BAC = BDCBAE = CDFCAF = BDE |
The conclusion is valid even when points B and C coalesce whereas line BC is replaced by the tangent to the circle. With this in mind, the problem appears to generalize the one of an accidental angle bisector. Also, in this formulation, the requirement of convexity is quite spurious.
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Suffice it to show that ADF + AEF = 180o. This is so because
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References
- T. Andreescu, R. Gelca, Mathematical Olympiad Challenges, Birkhäuser, 2004.
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