1999 USAMO Problems/Problem 6


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.


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

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

See Also

1999 USAMO (ProblemsResources)
Preceded by
Problem 5
Followed by
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