The incircle ω of triangle ABC touches its sides BC,CA and AB at points D,E and E, respectively. Point G lies on circle ω in such a way that FG is a diameter. Lines EG and FD intersect at point H. Prove that AB∥CH.
Problem
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Tags: geometry, parallel, incircle, Kharkiv
11.03.2020 05:11
parmenides51 wrote: The incircle ω of triangle ABC touches its sides BC,CA and AB at points D,E and F, respectively. Point G lies on circle ω in such a way that FG is a diameter. Lines EG and FD intersect at point H. Prove that AB∥CH. FTFY?
11.03.2020 05:28
11.03.2020 06:26
config note: C is the circumcenter of △EDH.
11.03.2020 18:47
Another solution is to use @Kagebaka's note and some simple steps of angle transformation to prove ∠BCM=∠ABC
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
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