# 1998 APMO Problems/Problem 4

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## Problem

(Răzvan Gelca) Let $ABC$ be a triangle and $D$ the foot of the altitude from $A$. Let $E$ and $F$ be on a line through $D$ such that $AE$ is perpendicular to $BE$, $AF$ is perpendicular to $CF$, and $E$ and $F$ are different from $D$. Let $M$ and $N$ be the midpoints of the line segments $BC$ and $EF$, respectively. Prove that $AN$ is perpendicular to $NM$.

## Solution

We use directed angles mod $\pi$.

$[asy] size(200); defaultpen(1); pair B=(0,0), A=(1,3), C=(4,0), X=(4,4); pair D=foot(A,B,C); path O1=circumcircle(A,B,D), O2=circumcircle(A,C,D); pair E=IntersectionPoint(O1,D--X,1), F=IntersectionPoint(O2,D--X,0); pair M=(B+C)/2; pair N=(E+F)/2; draw(C--F--A--B--C--A); draw(B--E--A--D--E--F); draw(A--N--M--A,dotted); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,S); label("E",E,SSE); label("F",F,E); label("M",M,S); label("N",N,ESE); [/asy]$

Since $\angle ADB$ and $\angle AEB$ are both right angles, points $ABDE$ are concyclic. It follows that $$\angle CBA \equiv \angle DBA \equiv \angle DEA \equiv \angle FEA .$$ Similarly, the quadrilateral $ACDF$ is cyclic, so $$\angle ACB \equiv \angle ACD \equiv \angle AFD \equiv \angle AFE.$$ Thus $ACB$ and $AFE$ are similar triangles, so $AFEN$ and $ACBM$ are similar figures. It follows that $$\angle AND \equiv \angle ANE \equiv \angle AMB \equiv \angle AMD,$$ so points $AMND$ are concyclic. Since $ADM$ is a right angle, it then follows that $ANM$ is a right angle, as desired. $\blacksquare$