Hello ordinalhigh

sa-ri-ga-ma has given you most of the proof, but if I may, I'd like to make one or two corrections and additions.

First, note that $\displaystyle E$ is the foot of the perpendicular from $\displaystyle B$ to $\displaystyle AC$, and $\displaystyle F$ the foot of the perpendicular from $\displaystyle C$ to $\displaystyle AB$. (I think sa-ri-ga-ma has these points the wrong way round.) Suppose $\displaystyle O$ is the orthocentre. (i.e the point where $\displaystyle AD, BE$ and $\displaystyle CF$ meet).

Then:$\displaystyle \angle BFC = \angle BEC = 90^o$ (given)

$\displaystyle \therefore BFEC$ is a cyclic quadrilateral (angles in same segment)

$\displaystyle \therefore \angle BEF = \angle BCF$ (same segment)

Also:$\displaystyle \angle ODC = \angle OEC = 90^o$ (given)

$\displaystyle \therefore OECD$ is a cylic quadrilateral (opposite angles supplementary)

$\displaystyle \angle BCF = \angle DCO$ (same angle)

But $\displaystyle \angle DCO = \angle DEO$ (same seg)$\displaystyle =\angle DEB$ (same angle)

But $\displaystyle \angle BCF = \angle BEF$ (proven)

$\displaystyle \therefore \angle DEB = \angle BEF$

Now $\displaystyle D_2$ is the reflection of $\displaystyle D$ in $\displaystyle AC$, and $\displaystyle E$ lies on $\displaystyle AC$.$\displaystyle \therefore DE = ED_2$

$\displaystyle \therefore \angle ED_2D = \angle EDD_2$ (isosceles $\displaystyle \triangle EDD_2$)$\displaystyle =\angle DEB$ (alternate angles, $\displaystyle BE \parallel DD_2$)

$\displaystyle =\angle BEF$ (proven)

$\displaystyle \therefore D_2EF$ is a straight line (corresponding angles equal, $\displaystyle BE \parallel DD_2$)

Similarly for $\displaystyle D_1FE$.

Grandad