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Saudi Arabian Mathematical Competitions

Geometry Difficulty 4.5 AIME Prove it Saudi Arabia

Let $ABCD$ be a rectangle of center $O$ , such that $\widehat{DAC}=60^{\circ}$ . The angle bisector of $\widehat{DAC}$ 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.

Solution

We have $\widehat{SAC}=\widehat{SCA}=30^{\circ}$ , so $SA=SC$ . Since $OA=OC$ , we get $SO \perp AC$ , hence $LA=LC$ . Moreover $\widehat{LAC}=60^{\circ}$ , so triangle $ALC$ is equilateral.

Point $S$ is the centroid of $\triangle LAC$ , thus $\frac{LS}{SO}=2$ . $DBCL$ is a parallelogram, hence $DQ=QC$ . We get that $M$ is the centroid of $\triangle DBC$ , so $\frac{CM}{MO}=2$ .

Finally, since $\frac{LS}{SO}=\frac{CM}{MO}$ , we obtain that $SM \parallel CL$ .

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