1998 APMO Problems/Problem 4

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. $\blacksquare$