# Proving that internal line segments are concurrent by Ceva's theorem for an incircle

I have the following question as follows.

The non-isosceles triangle $$\Delta ACE$$ below has incircle $$C(O,r)$$. The line segments AC, CE and EA are tangent to $$C(O,r)$$ at points $$B$$, $$D$$ and $$F$$ respectively. Prove that $$AD$$, $$CF$$ and $$EB$$ are concurrent.

I understand the problem and I can prove that the ANGLE BISECTORS are concurrent because if I let $$l_1$$, $$l_2$$, $$l_3$$ be the angle bisectors, then I have $$b'$$, $$a'$$ and $$c'$$ be the non vertex points of the angle bisectors of $$l_1$$, $$l_2$$, $$l_3$$ respectively. So by Ceva's theorem + the angle bisector theorem, we get:

$$\frac{Ea'}{a'C}\frac{Cb'}{b'a}\frac{Ac'}{c'E}=\frac{EA}{AC}\frac{CE}{EA}\frac{AC}{CE}=1$$. So this means the cevians (or angle bisectors here) must be parallel or concurrent but internal cevians can never be parallel so they must be concurrent. So the angle bisectors are concurrent.

I'm not sure how to use this information to prove that $$AD$$,$$CF$$ and $$EB$$ are concurrent though. I can physically see it since it's concurrent but I can't figure out how to relate the information I know to what I need. Can someone help me out?

• Their concurrence is the existence theorem of the so-called Gergonne point. See for that slide 25 of this presentation. Commented Nov 4, 2021 at 23:22

Tangent segments from an outside point to a circle are equal in length. So we have $$AB=AF$$, $$CB=CD$$ and $$ED=EF$$. Now the proof should be straightforward.