The incircle $\omega$ of triangle $ABC$ touches its sides $BC, CA$ and $AB$ at points $D, E$ and $E$, 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 $AB \parallel CH$.
Problem
Source:
Tags: geometry, parallel, incircle, Kharkiv
BestChoice123
11.03.2020 05:11
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?
MP8148
11.03.2020 05:28
Let $I = \overline{EF} \cap \overline{DG}$. By Pascal's on $DDFEEG$ we know that $C \in \overline{HI}$. By Pascal's on $FFDGGE$ we know that $\overline{AB} \parallel \overline{HI}$. Done.
Kagebaka
11.03.2020 06:26
config note: $C$ is the circumcenter of $\triangle EDH.$
wassupbros
11.03.2020 18:47
Another solution is to use @Kagebaka's note and some simple steps of angle transformation to prove $\angle BCM=\angle ABC$
jayme
12.08.2020 18:42
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
jayme
20.08.2020 12:08
Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/08.%201.%20Parallele%20a%20un%20cote%20du%20triangle.pdf p. 5-6. Sincerely Jean-Louis