Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.5 AIME Prove it India

In an acute triangle $ABC$ , $AD$ is the median, $BE$ is the internal angle bisector and $CF$ is the altitude, with $D$ , $E$ , $F$ respectively on sides $BC$ , $CA$ , $AB$ . If $DEF$ is equilateral, prove that $ABC$ is also equilateral.

Solution

Observe that in the right triangle $BFC$ , $BD = DC = a/2$ . Hence $DF = a/2$ and this gives $DE = EF = a/2$ . Now in triangle $BEC$ , $D$ is the midpoint of $BC$ and $DE = DB = DC$ . Hence $\angle BEC = 90^\circ$ . Since $BE$ bisects $\angle ABC$ , we conclude that $BA = BC$ and $CE = EA$ .

In the right triangle $CFA$ , $E$ is the midpoint of $AC$ . Hence $EC = EA = EF = a/2$ . We conclude that $CA = a = CB$ .

We obtain $AB = BC = CA$ . Hence $ABC$ is equilateral.

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