Let $ABCD$ be a quadrilateral (with non-perpendicular diagonals). The perpendicular from $A$ to $BC$ meets $CD$ at $K$. The perpendicular from $A$ to $CD$ meets $BC$ at $L$. The perpendicular from $C$ to $AB$ meets $AD$ at $M$. The perpendicular from $C$ to $AD$ meets $AB$ at $N$. 1. Prove that $KL$ is parallel to $MN$. 2. Prove that $KLMN$ is a parallelogram if $ABCD$ is cyclic.
Problem
Source: 2015 Pan-African Mathematics Olympiad Problem 6
Tags: geometry, parallelogram, Parallel Lines
LREDK
26.08.2015 19:09
not easy，may need a lot time
LeVietAn
26.08.2015 20:05
My solution: a) Easy to get $A$ be the orthocenter of triangles $CKL$ and $CMN$ $\Rightarrow AC\perp KL$ and $AC\perp MN$ $\Rightarrow KL\parallel MN$. b) Let $O_1, O_2$ be the circumcenter of triangles $CKL$ and $CMN$, resp. Let $X, Y$ be the midpoints of $KL, MN$, resp. We have $O_1X\overset{\parallel }{=}2CA$ (Well-know)and $O_2Y\overset{\parallel }{=}2CA$ $(*)$ On the other hand, $\measuredangle KO_1L=2\measuredangle KCL$(Well-know) $=2\measuredangle BCD=2(180^{\circ }-\measuredangle BAD)=2\angle MCN$($\because CM\perp AB,CN\perp AD$)$=\measuredangle NO_2M$ $(**)$ Since $(*), (**)$ deduced $\triangle O_1KL=\triangle O_2NM$ $\Rightarrow O_1K=O_2N$ $\Leftrightarrow 2O_1C=2O_2C$ $\Leftrightarrow \frac{KL}{sin\measuredangle KCL}=\frac{MN}{sin\measuredangle MCN}$ $\Leftrightarrow KL=MN$($\because \measuredangle KCL=\measuredangle NCM$), combined with $KL\parallel MN$(part a)$\Rightarrow KLMN$ is a parallelogram. DONE
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