clovnerie0q

2022-10-08

An Olympiad Geometry problem with incenter configurations.
Let the inscribed circle of triangle ABC touches side BC at D ,side CA at E and side AB at F. Let G be the foot of the perpendicular from D to EF. Show that $\frac{FG}{EG}=\frac{BF}{CE}$.
So this problem is equivalent to proving similarity between $\mathrm{\Delta }BFG$ and $\mathrm{\Delta }CEG$.
I was able to prove that $\mathrm{\angle }GFB$= $\mathrm{\angle }GEC$ but after that, I hit a dead end. I found the point G pretty annoying as I couldn't apply any circle theorems to it.

vakleraarrc

Explanation:
Just direct angle chasing and trigo will help you express those ratios.
- Show that $BF/CE=BD/DC=\left(BD/DG\right)/\left(DC/DG\right)=\mathrm{tan}\left({90}^{\circ }-\beta /2\right)/\mathrm{tan}\left({90}^{\circ }-\gamma /2\right)=\mathrm{tan}\gamma /2/\mathrm{tan}\beta /2$
- Show that $\mathrm{\angle }IFE=\mathrm{\angle }IAE=\alpha /2$.
- Show that $FG/GE=\left(FG/GD\right)/\left(GE/GD\right)=\mathrm{tan}\gamma /2/\mathrm{tan}\beta /2$.

st3he1d0t