My student showed me the following question.
A circle inscribed in triangle $ABC$ touches $AB$ at $D$, $BC$ at $E$, $AC$ at $F$. $G$ is on $DE$ such that $FG\perp DE$. Prove $FG$ bisects $\angle{AGC}$.
I managed to prove this using cartesian coordinates, but my proof takes about two pages of equations. I'm looking for a more elegant proof.
(Here is an outline of my proof. Let the circle have cartesian equation $x^2+y^2=1$ and assign coordinates: $D\space (-p,-\sqrt{1-p^2})$, $E\space (p,-\sqrt{1-p^2})$, $F\space (-q,\sqrt{1-q^2})$. In terms of $p$ and $q$, find the equations of the lines that make the three sides of the triangle, then find the coordinates of $A$, $C$ and $G$. Then show that the sum of gradients of $AG$ and $CG$ is $0$, so $FG$ bisects $\angle{AGC}$.)
Best Answer
Let $O$ be the center of the incircle of $\triangle ABC$ and the circumcircle of $\triangle DEF$. Hence $OF\perp AC$, $\angle OAF=\frac{\angle A}{2}$ and $\angle OCF=\frac{180^{\circ}-\angle A-\angle B}{2}.$
Since $AD, AF$ are tangents and using Alternate Segment Theorem, $$AD=AF=a~~\text{and}~~\angle ADF=\angle AFD=\angle FED=90-\frac{\angle{A}}{{2}}.$$ Using the same argument for the other tangents, it is easy to see that $$CE=CF=c~~\text{and}~~\angle FEC=\angle EDF=\frac{\angle A+\angle B}{2}\implies \angle ADG=CEG.$$ Using the above facts, $$\triangle AOF\sim \triangle FEG~~\text{and} ~~\triangle COF\sim\triangle FDG\implies \frac{a}{DG}=\frac{c}{GE}.$$ Therefore, $$\triangle ADG\sim \triangle CEG\implies \angle CGE=\angle AGD\implies \angle AGF=\angle CGF.$$