# Prove that in an obtuse triangle the orthocentre is the excenter of the orthic triangle

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?

• What are $D'$ and $F'$? – S.B. Jul 15 '13 at 16:59
• "Extend $ED$ to $D'$ and $EF$ to $F'$" – Gerard Jul 15 '13 at 17:00
• That's why I asked. Extend to where? – S.B. Jul 15 '13 at 17:02
• To wherever you like :P. It doesn't matter. Its just for naming the angles. – Gerard Jul 15 '13 at 17:02
• I see $D'$ and $F'$ aren't particularly significant. This is a basic result about orthocenter (it also applies for acute triangles though $H$ would correspond to the incenter of $FED$). Just use the fact that HDEA and AFDC are cyclic. – S.B. Jul 15 '13 at 17:11

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