AoPS - Problem

Source: Turkey JBMO TST 2016 P5

Tags: geometry proposed, perpendicular lines, geometry

crazyfehmy

23.05.2016 01:18

In an acute triangle $ABC$, the feet of the perpendiculars from $A$ and $C$ to the opposite sides are $D$ and $E$, respectively. The line passing through $E$ and parallel to $BC$ intersects $AC$ at $F$, the line passing through $D$ and parallel to $AB$ intersects $AC$ at $G$. The feet of the perpendiculars from $F$ to $DG$ and $GE$ are $K$ and $L$, respectively. $KL$ intersects $ED$ at $M$. Prove that $FM \perp ED$.

Notice that $\angle FME= \angle 90$ means $LFME$ should be cyclic.Let $\angle ECB = \angle X$ and $\angle DAC = \angle Y$ . $\angle ECB= \angle FEC=\angle X$ ,and $\angle DAC = \angle DEC= \angle Y$ . So $\angle FEM = \angle (X+Y)$ . If $\angle FLM = \angle (X+Y)$ we will be done.Notice that $\angle BAC = \angle FGD=\angle (X+Y)$ since $AB$ is parallel to $GD$. This means that if $LKGF$ is cyclic, then $\angle FME = \angle 90$ . And since $\angle GKF= \angle GLF= \angle 90$ , $LKGF$ is cyclic, and we're done !