SM and CL are parallel

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SM and CL are parallel

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Source: Romanian MO 2010 Grade 7

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#1 h Aug 6, 2012, 12:26 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Let $ABCD$ be a rectangle of centre $O$ , such that $\angle DAC=60^{\circ}$ . The angle bisector of $\angle DAC$ meets $DC$ at $S$ . Lines $OS$ and $AD$ meet at $L$ , and lines $BL$ and $AC$ meet at $M$ . Prove that lines $SM$ and $CL$ are parallel.

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#2 h Aug 7, 2012, 1:42 AM • 2 Y

Y by Adventure10, Mango247

Let $AD=x$ . Thus $AB=x\sqrt3$ . Note that $ASD$ and $LSD$ are congruent, so $AD=LD$ , $ACL$ is equilateral and $AL=2x$ .
$\frac{AM}{sin\angle ALM}=\frac{ML}{sin60^\circ} \implies ML=\frac{\sqrt3}{2}\frac{AM}{sin\angle ALM}=\frac{\sqrt7}{2}AM \implies BM=x\sqrt7 - \frac{\sqrt7}{2}AM$ .
Stewart's Theorem in $ABL$ gives $x\sqrt7(AM^2+\frac{\sqrt7}{2}AM(x\sqrt7-\frac{\sqrt7}{2}AM))=3x^2\frac{\sqrt7}{2}AM+4x^2(x\sqrt7-\frac{\sqrt7}{2}AM) \implies AM=2x, \frac{4x}{3}$ .
$AC=2x$ so $AM=\frac{4x}{3}$ . From $ADS$ , $AS=\frac{2x}{\sqrt3}$ . Thus from $ASM$ it follows that $\angle ASM=90^\circ$ so $\angle AMS=60^\circ=\angle ACL$ and the desired conclusion follows.

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#3 h Aug 7, 2012, 7:21 AM • 2 Y

Y by Adventure10 and 1 other user

An outline of my solution:
Obviously, $\triangle ACS$ is isosceles and $\triangle ACL$ equilateral. Take ${\{P|}\in BC\cap OL$ . thus $S$ is the centroid of $\triangle ACL$ , whilst $M$ the centroid of $\triangle CLP\implies\frac{CS}{SD}=\frac{CM}{MO}=2\iff MS\parallel DO$ , but $DO\parallel CL$ , done.

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#4 h Aug 9, 2012, 2:24 AM • 2 Y

Y by Adventure10, Mango247

We know $\triangle ACL$ is equilateral; also, from ratios, $3DS = CD \Longrightarrow \frac{CS}{CD} = \frac{2}{3}$ . Now $\triangle BMC \sim \triangle LMA$ . Since $AL = 2BC$ , $AM = 2CM$ . So $\frac{CM}{CD} = \frac{2}{3}$ . By SAS, $\triangle LMS \sim \triangle LOD$ , so $SM \parallel DO$ . Since $DO \parallel CL$ , we're done.

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