IMO Hk TST

Geometry Difficulty 7.8 National Olympiad, round 2 Prove it Hong Kong

Let $ABCDEF$ be a convex hexagon such that $AB \parallel DE$ , $BC \parallel EF$ and $CD \parallel FA$ . Let $H$ , $I$ , $J$ , $K$ , $L$ and $M$ be the midpoints of sides $AB$ , $BC$ , $CD$ , $DE$ , $EF$ and $FA$ , respectively. Prove that lines $HK$ , $IL$ and $JM$ are concurrent.

Solution

Let $AD$ meet $BE$ at $X$ , $BE$ meet $CF$ at $Y$ , and $CF$ meet $AD$ at $Z$ . Since $AB \parallel ED$ , we know that $\triangle XAB \sim \triangle XDE$ . Also, the midpoints $H$ and $K$ of $AB$ and $DE$ are corresponding points under this similarity. Thus, $X$ lies on $HK$ . Similarly, $Y$ lies on $IL$ , and $Z$ lies on $JM$ .

By Ceva's theorem, it suffices to prove
$\frac{\sin \angle AXH}{\sin \angle HXB} \times \frac{\sin \angle EYL}{\sin \angle LYF} \times \frac{\sin \angle CZJ}{\sin \angle JZD} = 1. \qquad (1)$
Firstly, since $AH = HB$ and $\triangle XAB \sim \triangle XDE$ , we have
$\frac{\sin \angle AXH}{\sin \angle HXB} = \frac{XB}{XA} = \frac{BE}{AD}.$
By symmetry, we have $\frac{\sin \angle EYL}{\sin \angle LYF} = \frac{CF}{BE}$ and $\frac{\sin \angle CZJ}{\sin \angle JZD} = \frac{AD}{CF}$ . Therefore, the left-hand side of (1) is
$\frac{BE}{AD} \times \frac{CF}{BE} \times \frac{AD}{CF} = 1$
as desired.

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