[Math] Prove that in an obtuse triangle the orthocentre is the excenter of the orthic triangle

Question

Consider an obtuse angled $\Delta ABC$ with altitudes $AD, BE, CF$ concurrent at $H$. Consider the orthic triangle $\Delta FED$. Extend $ED$ to $D'$ and $EF$ to $F'$. Prove that $\angle FDH = \angle HDD'$ and $\angle DFH = \angle HFF'$. In other words prove that $H$ is the excenter of $\Delta FED$.

I tackled $\angle FDH = \angle HDD'$ first.
I tried reducing the proof to a simpler statement:

Since $\angle D'DH = \angle ADE$, and $\angle FDH = 90 – \angle FDB$, it is sufficient to prove that $\angle ADE + \angle FDB = 90$

Now, the proof hinges on the conjecture that in an orthic triangle of an obtuse triangle, the point with the obtuse angle is the incenter of the orthic triangle. I was unable to prove this conjecture. Is there a proof for this conjecture (or is it incorrect altogether?), or is there an alternative proof to the whole problem?

Answer 1

Hint:

• Consider angles inscribed in circumcircles of the following quadrilaterals: $BDHF$, $ADBE$ and $BFCE$.

I hope this helps ;-)