The diagram looks simple, but I can't really think of a penultimate step. I also tried bashing it but failed. Can someone hint me at the right direction?

$\angle BFC=45^0$ and $\angle AEF = \arctan 2$ Easy to fing $\tan \angle FHE=3$

Or solution without using trigonometry. Let $AB$ intersect $CF$ in $G$ and $BF$ intersect $GD$ in $M$ $AG=CD$ and $\angle ABF=\angle ACF$ so easy to prove ,that $GM=MD$ $\to$ $AD$ and $BM$ are medians. So $AD=2AH$

claim $BF$ bisects $AO$. indeed $\frac{BA}{BO}.\frac{\sin \angle ABF}{\sin \angle FBD}=\frac{BA}{BO}.\frac{\sin \angle ACF}{\sin \angle FCD}=\frac{BA}{BO}.\frac{DC}{AC}=1$ . consider $H'$ the midpoint of $DH$ then $OH'\parallel BH$ so $BF$ bisects also $ AH'$ . RH HAS

Attachments:

Dear Mathlinkers, 1. T the point of intersection of AB and CE ; A is the midpoint of TB 2. TD is tangent to (O) at D (by the converse of the angle tangente theorem) 3. THO is the Pascal's line of Pascal DDACFBD 4. H is the median point of the triangle DTB and we are done... Sincerely Jean-Louis

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20V.pdf p. 47... Sincerely Jean-Louis