Problem 2 of the Australian Mathematical Olympiad 2008

Let ABC be an acute triangle, and let D be the point on AB (extended if necessary) such that AB and CD are perpendicular. Further, let tA and tB be the tangents to the circumcircle of ABC through A and B, respectively, and let E and F be the points on tA and tB, respectively, such that CE is perpendicular to tA and CF is perpendicular to tB. Prove that CD/CE = CF/CD.

Solution by Steve Dinh, a.k.a. V Đức Din (dedicated to Lisa Nguyen of CapGroup)

We see that ADCE and BDFC are both concyclic which cause

∠DFC = ∠ABC (*), ∠DEC = ∠BAC (**) and

∠BAE + ∠DCE = ∠ABF + ∠DCF = 180 but ∠BAE = ∠ABF (both subtend larger arc AB); therefore, ∠DCE = ∠DCF.

However, ∠ACB subtends smaller arc AB; hence ∠ACB = ∠DCE = ∠DCF since ∠ACB also combines with ∠BAE to be 180.

Now with the addition of (*), the triangles ABC and DFC are similar because their respective angles are equal. Similarly, combined with (**), the two triangles ABC and EDC are also similar for the same reason. Therefore, triangles DFC and EDC are similar, and CD/CE = CF/CD.

Remark: It is a curious observation that this problem made its appearance at different times and in a little different guises in three mathematical olympiads. Besides the 2008 Australian Olympiad, it showed up also at the the 2010 Canadian Mathematical Olympiad and even much earlier in a published form.