The orthic triangle

Much can be learned from inspection of Figure 1, which shows an acute-triangle ABC, its orthocenter H (the common point of the altitudes), and its orthic triangle DEF. (Using the same notations as in the previous lecture, here AD, BE and CF are the altitudes of $\Delta ABC$.)

  

Figure 1: The orthic triangle

Problem 1. H is the incenter of $\Delta DEF$. In other words, the orthocenter of an acute triangle is the incenter of its orthic triangle.

Proof. The right triangles ABE and ACF having common angle A give us the same value for $\angle EBA$ and $\angle ACF$. The equality of these last two angles could also have been seen from the fact that, since $\angle BEC$ and $\angle BFC$ are right angles, the quadrilateral BCEF is inscribable in a circle. Making analogous use of the quadrilaterals BDHF and CEHD, we find that

$$\angle HDF = \angle HBF = \angle EBF = \angle ECF = \angle ECH = \angle EDH.$$

Thus HD bisects $\angle EDF$. Similarly HE bisects $\angle FED$, and HF bisects $\angle DFE$.

Problem 2. Prove that $\Delta AEF \sim \Delta DBF \sim \Delta DEC \sim \Delta ABC$.

Hint. To prove that $\Delta AEF \sim \Delta ABC$ use the fact that the quadrilateral BCEF is inscribable in a circle, so $\angle B=\angle FEA$.