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Problem 1189

National olympiad, first round

Geometry Difficulty 6.2 Prove it Spring Mathematical Tournament · Bulgaria

Given a $\triangle ABC$ . A circle $k$ through $A$ and $B$ intersects the sides $AC$ and $BC$ at points $L$ and $N$ , respectively. Let $M$ be the midpoint of the arc $LN$ lying in the triangle. Set $AM \cap BL = D$ , $AM \cap BN = F$ , $BM \cap AL = G$ and $BM \cap AN = E$ . Prove that:

a) $DE\parallel FG$ ;

b) if $DEFG$ is a parallelogram, it is a rhombus.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Let $AN \cap BL = P$ .

a) Since $\angle LAM = \angle MAN = \angle LBM = \angle MBN$ , the quadrilaterals $ABED$ and $ABFG$ are cyclic. Then $\angle AED = \angle ABL = \angle ANL$ and hence $DE\parallel LN$ . Analogously $FG\parallel LN$ .

b) Since $\frac{DP}{LP} = \frac{DE}{LN} = \frac{GF}{LN} = \frac{CF}{CN}$ , then $\frac{LD}{DP} = \frac{NF}{FC}$ . Hence
$\frac{LA}{AP} = \frac{LD}{DP} = \frac{NF}{FC} = \frac{NA}{AC}.$
It follows that $\triangle APL \sim \triangle ACN$ which gives $\angle APL = \angle ACB$ , i.e. $LPNC$ is a cyclic quadrilateral. Then
$\begin{align*} 180^\circ &= \angle APB + \angle ACB = 180^\circ - \angle PAB - \angle PBA + 180^\circ - \angle CAB - \angle CBA \\ &= 2 \cdot 180^\circ - (\angle PAB + \angle CAB) - (\angle PBA + \angle CBA) \\ &= 2(180^\circ - \angle MAB - \angle MBA) = 2 \angle AMB, \end{align*}$
i.e. $\angle AMB = 90^\circ$ . So $DF \perp EG$ and therefore the parallelogram $DEFG$ is a rhombus.

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