USAMO 1999/6

For discussing Olympiad level Geometry Problems
tanmoy
Posts:312
Joined:Fri Oct 18, 2013 11:56 pm
Location:Rangpur,Bangladesh
USAMO 1999/6

Unread post by tanmoy » Wed Feb 25, 2015 12:05 am

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"

Tahmid
Posts:110
Joined:Wed Mar 20, 2013 10:50 pm

Re: USAMO 1999/6

Unread post by Tahmid » Wed Feb 25, 2015 10:55 am

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 $DC$

tanmoy
Posts:312
Joined:Fri Oct 18, 2013 11:56 pm
Location:Rangpur,Bangladesh

Re: USAMO 1999/6

Unread post by tanmoy » Wed Feb 25, 2015 1:15 pm

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 $DC$
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 $A$.
"Questions we can't answer are far better than answers we can't question"

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