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.

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

Beginner2022-10-09Added 6 answers

Explanation:

Just direct angle chasing and trigo will help you express those ratios.

- Show that $BF/CE=BD/DC=(BD/DG)/(DC/DG)=\mathrm{tan}({90}^{\circ}-\beta /2)/\mathrm{tan}({90}^{\circ}-\gamma /2)=\mathrm{tan}\gamma /2/\mathrm{tan}\beta /2$

- Show that $\mathrm{\angle}IFE=\mathrm{\angle}IAE=\alpha /2$.

- Show that $FG/GE=(FG/GD)/(GE/GD)=\mathrm{tan}\gamma /2/\mathrm{tan}\beta /2$.

Just direct angle chasing and trigo will help you express those ratios.

- Show that $BF/CE=BD/DC=(BD/DG)/(DC/DG)=\mathrm{tan}({90}^{\circ}-\beta /2)/\mathrm{tan}({90}^{\circ}-\gamma /2)=\mathrm{tan}\gamma /2/\mathrm{tan}\beta /2$

- Show that $\mathrm{\angle}IFE=\mathrm{\angle}IAE=\alpha /2$.

- Show that $FG/GE=(FG/GD)/(GE/GD)=\mathrm{tan}\gamma /2/\mathrm{tan}\beta /2$.

st3he1d0t

Beginner2022-10-10Added 3 answers

Step 1

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