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Irish Mathematical Olympiad, 2014

Geometry Difficulty 6.3 National Olympiad Prove it Ireland

Let $ABC$ be a triangle with greatest side $BC$ and let $D$ and $E$ be internal points on $AB$ and $AC$ , respectively, such that $|AD| = |AE|$ . Let $F$ and $G$ be internal points on $BC$ such that $|BD| = |BG|$ and $|CF| = |CE|$ . Prove that $D$ , $E$ , $F$ and $G$ are concyclic.

Solution

The relative position of $F$ and $G$ leads to different situations. If $F = G$ we have nothing to show. There remain two cases to consider: case (i) $F$ is between $C$ and $G$ ; case (ii) $G$ is between $C$ and $F$ .

case (i)

case (ii)
In both cases, denote $\alpha = \angle ADE = \angle AED$ , $\beta = \angle BDG = \angle BGD$ and $\gamma = \angle CFE = \angle CEF$ . Considering the angle sum of the three triangles involving these angles, we obtain $2(\alpha + \beta + \gamma) = 540^\circ - (\angle A + \angle B + \angle C) = 360^\circ$ , hence $\alpha + \beta + \gamma = 180^\circ$ .
In case (i) we have $\angle EDG + \angle EFG = 180^\circ - (\alpha + \beta) + 180^\circ - \gamma = 180^\circ$ , hence $DEFG$ is a cyclic quadrilateral.
In case (ii) we have $\angle EDG = 180^\circ - (\alpha + \beta) = \gamma = \angle EFG$ , which implies that $EDFG$ is a cyclic quadrilateral.

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