**Problem 6 of USA Mathematical Olympiad 1999**

Let ABCD be an isosceles trapezoid with AB || CD. The inscribed circle w of triangle BCD meets CD at E. Let F be a point on the (internal) angle bisector of ∠DAC such that EF ⊥ CD. Let the circumscribed circle of triangle ACF meet line CD at C and G. Prove that the triangle AFG is isosceles.

**Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to my former colleague Lo Ngo)**

Let w be the circumcircle of triangle ACF. Draw the incircle of ΔADC with center at I’; this circle is symmetrical of the incircle of ΔBDC via the axis passing through centers of AB and DC.

Draw the circumcircle w_{1} of ΔADC to intercept AF at M. Since AF is the
bisector of ∠DAC, we have MD = MI, and M is the center of circle w_{2} as shown.
From M draw a line ⊥ to DC and meets it at N. Since N is the midpoint
of EE’, MI’ = MF and therefore F is on circle w_{2}.

For circle w_{1}, we have: AP × PM = DP × PC (1)
For circle w, we have: AP × PF = GP × PC (2)

From (1) and (2) PM / PF = DP / GP (3)

or MD || GF and

MD / GF = PM / PF

or GF = MD × PF / PM (4)

For circle w_{2}, we have: IP × PF = DP × PC (5)

From (5) and (2), we have: IP / AP = DP / GP (6)

From (6) and (3), we have: IP / AP = PM / PF (7)

From (7) IP / AP = PM / PF = (IP+PM) / (AP+PF) = IM / AF (8)

From (4) and (8) GF = MD × AF / MI = AF

Therefore, ΔAFG is isosceles.