### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

Explanation 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 = ∠EDF and, in particular, ∠CAF = ∠BDE.

In a slightly different form the problem appeared in [Andreescu & Gelca, p. 9, problem 1.2.7]:

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 = ∠BDC and ∠BAE = ∠CDF. Prove that ∠CAF = ∠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.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

Suffice it to show that ∠ADF + ∠AEF = 180°. This is so because

 ∠ADF + ∠AEF = (∠ADC + ∠CDF) + (∠ABF - ∠BAE) = (∠ADC + ∠ABC) + (∠CDF - ∠BAE) = ∠ADC + ∠ABC = 180°.

### References

1. T. Andreescu, R. Gelca, Mathematical Olympiad Challenges, Birkhäuser, 2004. ### Chasing Inscribed Angles 