USAMO 1999/6
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.
"Questions we can't answer are far better than answers we can't question"
Re: USAMO 1999/6
tanmoy ,
did you solve this ? ...
if not , then try to show that $E$ is the touch point of the excircle of triangle $ADC$ with side $DC$
did you solve this ? ...
if not , then try to show that $E$ is the touch point of the excircle of triangle $ADC$ with side $DC$
Re: USAMO 1999/6
Yeah,I have solved this.Actually the main part of the problem is to proof that $F$ is the center of the excircle of triangle $ADC$ opposite $A$.Tahmid wrote:tanmoy ,
did you solve this ? ...
if not , then try to show that $E$ is the touch point of the excircle of triangle $ADC$ with side $DC$
"Questions we can't answer are far better than answers we can't question"