# Thread: Orthic triangle problem

1. ## Orthic triangle problem

Suppose that triangle ABC is an acute-angle triangle, D is the foot of the altitude from A and D1 and D2 are the reflections of D with respect to AB and AC, respectively. Let F and E be the feet of the altitudes from C and B respectively. Show that the points F, E, D1, D2 are collinear.

I am thinking that if I can show that angle $DD_{1}E$ = angle $DD_{1}F$ = angle $DD_{1}D_{2}$ then that would show the points are collinear. I had a thought I can establish this using cyclic quadrilaterals but I got stuck. I can use some help on what to do next.

Thanks.

2. Not sure I have understood but if "angle = angle = angle " These points form sets of angles then it seems F, E, D1, D2 can not be collinear.

I may not have a correct understanding of "reflection" please clarify.

3. BEFC is a cyclic quadrilateral. Therefore

4. BEFC is a cyclic quadrilateral. Therefore
angle BFE = angle BCE
If O is the orthocenter, then DOFC is a cyclic quadrilateral. Therefore
angle OFD=angle OCD
Therefore
angle EFO=angle OFD
Now
BF is parallel to DD_2
Therefore
angle OFD=angle FDD_2 = angle FD_2D
Therefore
angle EFD=angle FDD_2 + angle FD_2D
Hence EFD_2 is a straight line.
Similarly you can show D_1EF also a straight line.

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

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

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

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

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

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

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

But $\angle DCO = \angle DEO$ (same seg)
$=\angle DEB$ (same angle)
But $\angle BCF = \angle BEF$ (proven)

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

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

$=\angle BEF$ (proven)

$\therefore D_2EF$ is a straight line (corresponding angles equal, $BE \parallel DD_2$)
Similarly for $D_1FE$.