1999 USAMO Problems/Problem 6

Problem

Let $ABCD$ be an isosceles trapezoid with $AB \parallel CD$ . The inscribed circle $\omega$ of triangle $BCD$ meets $CD$ at $E$ . Let $F$ be a point on the (internal) angle bisector of $\angle DAC$ such that $EF \perp CD$ . Let the circumscribed circle of triangle $ACF$ meet line $CD$ at $C$ and $G$ . Prove that the triangle $AFG$ is isosceles.

Solution

Quadrilateral $ABCD$ is cyclic since it is an isosceles trapezoid. $AD=BC$ . Triangle $ADC$ and triangle $BCD$ are reflections of each other with respect to diameter which is perpendicular to $AB$ . Let the incircle of triangle $ADC$ touch $DC$ at $K$ . The reflection implies that $DK=CE$ , which then implies that the excircle of triangle $ADC$ is tangent to $DC$ at $E$ . Since $EF$ is perpendicular to $DC$ which is tangent to the excircle, this implies that $EF$ passes through center of excircle of triangle $ADC$ .

We know that the center of the excircle lies on the angular bisector of $DAC$ and the perpendicular line from $DC$ to $E$ . This implies that $F$ is the center of the excircle.

Now $\angle GFA=\angle GCA=\angle DCA$ . $\angle ACF=90^\circ+\frac{\angle DCA}{2}$ . This means that $\angle AGF=90^\circ-\frac{\angle ACD}{2}$ . (due to cyclic quadilateral $ACFG$ as given). Now $\angle FAG =180^\circ- (\angle AFG + \angle FGA)=90^\circ-\frac{\angle ACD}{2}=\angle AGF$ .

Therefore $\angle FAG=\angle AGF$ . QED.

1999 USAMO (Problems • Resources)
Preceded by Problem 5	Followed by Last Question
1 • 2 • 3 • 4 • 5 • 6
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1999 USAMO Problems/Problem 6

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