Let $\triangle ABC$ be a triangle. Let $M$ be the midpoint of side $[BC]$. $H,$ and $I$ are respectively the orthocenter and incenter of $\triangle ABC$. Let $D = (MH)\cap(AI)$. $E$ and $F$ are the feet of perpendiculars from $D$ to $(AB)$ and $(AC)$, respectively. Prove that $E, F$ and $H$ are collinear.
Here is the source of the problem (in french) here
I have solved it using barycentric coordinates. As a matter of fact, one can get that lines $(MH)$ and $(AI)$ have equations, respectively: $\left[\displaystyle \frac{c^2-b^2} {S_{BC}}:\frac1{S_A}:-\frac1{S_A}\right]$ and $\left[\displaystyle 0:-c:b\right]$
$($Here, $S_A=\displaystyle \frac{b^2+c^2-a^2}2$, define cyclically $S_B$ and $S_C$, it's Conway's Notation)
Intersecting these lines gives un-normalized: $D\left(\displaystyle\frac{S_{BC}}{S_A(b+c)}:b:c\right)$,
which in turn gives: $F\left(\displaystyle\frac{S_C}{b}+\frac{S_{BC}}{S_A(b+c)}:0:\frac{S_A}{b}+c\right)$
and: $E\left(\displaystyle\frac{S_B}{c}+\frac{S_{BC}}{S_A(b+c)}:\frac{S_A}{c}+b:0\right)$.
Now, clearly the deteminant formed by $E,F$ and $H$ is null. The conclusion follows.
What I'm asking for is a synthetic solution to this problem. I have tried to come up with one, but couldn't. The main thing I noticed is the line connecting the two touch-points of the incircle with sides $(AC)$ and $(AB)$ is parallel to line $(EF)$, so maybe what we're looking for is a convenient homethety.
Best Answer
In a configuration involving the orthocenter, $H$, and the midpoint of a side, $M$, it is almost certainly a good idea to consider the circumcircle of the triangle. Because we have a pretty useful result concerning the line $HM$ and the circle $(ABC)$: $HM = MP$ and $A, O, P$ are collinear where $O$ is the circumcenter and $P$ is the intersection point of $HM$ and $(ABC)$ that lies on the other side of the line $BC$. Let $Q$ be the other intersection point. Since $\angle{AQP} = \angle{AQD} = \angle{AED} = \angle{AFD} = 90°,$ we know that points $A, Q, E, D, F$ all lie on a circle with diameter $AD$.
Let $\measuredangle{XYZ}$ denote the directed angle between lines $XY$ and $YZ$ modulo $180°$. The proposition $\measuredangle{AEX} + \measuredangle{AFX} = 0$ implies that $X$ lies either on $EF$ or $AI$. The problem statement makes it clear that $H$ is not on the line $AI$. Therefore, it suffices to show that $\measuredangle{AEH} + \measuredangle{AFH} = 0$.
Let's work backwards. How can we obtain such an equation? Note that $\measuredangle{AEH} = \measuredangle{BEH}$ and $\measuredangle{AFH} = \measuredangle{CFH}.$ Moreover, it is well-known that $\measuredangle{HBE} + \measuredangle{HCF} = \measuredangle{HBA} + \measuredangle{HCA} = 0.$ Thus, it looks like a promising strategy to show that $\triangle{HEB} \sim -\triangle{HFC}.$
Trying to obtain ratios involving the sides of these two triangles and chasing angles using the concyclicity condition we've proved above, it is easy to find out that $\triangle{QEB} \sim \triangle{QFC}$: $\measuredangle{QBE} = \measuredangle{QBA} = \measuredangle{QCA} = \measuredangle{QCF}$ and $\measuredangle{QEB} = \measuredangle{QEA} = \measuredangle{QFA} = \measuredangle{QFC}.$ Therefore, $\frac{QB}{QC} = \frac{EB}{FC}$. What remains to prove is that $\frac{QB}{QC} = \frac{HB}{HC}.$ It turns out that this is fairly simple to show as well: Since $BM = MC$, $\triangle{QBP}$ and $\triangle{QCP}$ have the same area. It implies that $QB \cdot PB = QC \cdot PC$. The fact that $HBPC$ is a parallelogram yields the desired result.
Now, we have all the required tools to write a completely synthetic proof. Hope this helps!