parmenides51 wrote: The incircle $\omega$ of triangle $ABC$ touches its sides $BC, CA$ and $AB$ at points $D, E$ and $F$, respectively. Point $G$ lies on circle $\omega$ in such a way that $FG$ is a diameter. Lines $EG$ and $FD$ intersect at point $H$. Prove that $AB \parallel CH$. FTFY?

Solution

config note: $C$ is the circumcenter of $\triangle EDH.$

Another solution is to use @Kagebaka's note and some simple steps of angle transformation to prove $\angle BCM=\angle ABC$

My two cents... See attachment.

Attachments:

Dear Mathlinkers, just to say that the Boutin's theorem leads to a collinearity with C and we finish with the Pascal's theorem... Sincerely Jean-Louis

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/08.%201.%20Parallele%20a%20un%20cote%20du%20triangle.pdf p. 5-6. Sincerely Jean-Louis