# USA 1999 Problem 6

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 w1 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 w2 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 w2.

For circle w1, 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 w2, 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.