Prove two line segments are perpendicular to each other in a triangle with orthocenter See below chart: 

$H$ is the orthocenter of $\triangle ABC$. $AD, BE$ are altitudes of the triangle. $F$ is on $AD$. $\angle BFD = \angle BAC$. $\frac{FH}{HD} = \frac{KF}{FD}$. $KL \perp BE$. Prove that $FL \perp AB$

My thoughts: obviously there are a lot of similar triangles..such as $\triangle BFD \sim \triangle BAE $,  $\triangle HKL \sim \triangle HAE \sim \triangle HBD$ etc. But $FL$ and $AB$ seem hard to relate. 

I think we can also extend $LK$ to intersect $AB$ at $J$, then we just need to prove that $JL^2 - JF^2 = BL^2 - BF^2$.. that might be closer to what we want to prove.
We will show that $ \angle FLH = \alpha \, ( = \angle BAC)$.
If so, then since $\angle LHC = \angle EHC = 90^ \circ - \angle HCE = \alpha$, so $FL \parallel HC \perp AB$.   



Let $ \angle FLH = \theta$.   

From the given condition, $ \frac{ FH}{ FK } = \frac{ HD } { FD } = \frac{ DH / BD } { FD / BD }  = \frac{ \tan ( 90^\circ - \gamma) } { \tan (90^ \circ - \alpha } = \frac{ \tan \alpha } { \tan \gamma} $.

By sine rule on $ \Delta FHL $, $\frac{ FH}{\sin \theta } = \frac{ FL } { \sin \gamma}$.
By sine rule on $\Delta KFL$, $ \frac{ KF} { \cos \theta } = \frac{ FL } { \cos \gamma}$.
Hence $ \frac{ FH } { FK } = \frac{ FH / FL } { KH / FL } = \frac{ \sin \theta / \sin \gamma } { \cos \theta / \cos \gamma } = \frac{ \tan \theta } { \tan \gamma}$.   

Comparing these equations, $ \tan \alpha = \tan \theta$. Since these angles are $ &lt; 180^\circ$, hence $ \theta = \alpha$.   



Other interesting results:
1.  $\Delta LHF \sim \Delta ABC$
2. Circumcircle of $LHF$ is tangential to $BF$

Note: I made the assumption that $\alpha &lt; 90 ^ \circ$ as hinted in the diagram. For completeness (in an olympiad), this proof could benefit from checking cases / using directed angles.
Experimental approach:

Extend LK to touch AB at Q.

In this optimized drawing we have:

$\frac{FH}{HD}≈ \frac{KF}{FD}$

$\angle FBD=\angle BFD=\angle BAC=\angle BQL=45^o$

Triangles HBD and HQL are similar so:

$\angle HBD=\angle HQL$

$\angle FQL +\angle BQF=45^o$

$\angle HBD +\angle FBH=45^o$

$\angle FBH=\angle BQF$

$\angle BQL=\angle QBL=45^0$

⇒$\angle FBQ=\angle FQB=\angle HBD$

Therefore LF must be a part of altitude of triangle QLB. that is LF is perpendicular to AB.