Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.7 AIME Prove it Brazil

Let $ABCD$ be a quadrilateral with $\angle ABC \neq 90^\circ$ . Let $M$ and $N$ be the midpoints of $AD$ and $CD$ , respectively. Prove that the lines perpendicular to $BC$ passing through $M$ and perpendicular to $AB$ passing through $N$ and $BD$ are concurrent if and only if the diagonals $BD$ and $AC$ are perpendicular.

Solution

Consider a homothety with center on $D$ that takes $M$ to $A$ and $N$ to $C$ . So the perpendicular lines are mapped to the altitudes of the triangle $ABC$ relative to $A$ and $C$ , and the intersection $P$ of the perpendicular lines is mapped to the orthocenter $H$ of triangle $ABC$ .

Notice that the condition that the perpendicular lines and $BD$ are concurrent is equivalent to $B$ , $P$ and $D$ being collinear. But the homothety implies that $D$ , $P$ and $H$ are collinear, so the three lines are concurrent if and only if $BH$ and $BD$ coincide, that is, $BD \perp AC$ , since $BH \perp AC$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.