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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.

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  • $\begingroup$ How is your problem a special case of China $1999$? Yufei Zhao's problem can "easily" be proved using a similar approach to the one employed to show the lemma. However, I cannot see how your problem should follow from that $\endgroup$
    – Dr. Mathva
    May 13 at 20:51
  • $\begingroup$ You are right! I somehow convinced myself yesterday that it should follow up from it, I'll edit it out! $\endgroup$
    – ZNatox
    May 13 at 21:08

3 Answers 3

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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!

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Let $T_1$ be the foot from $H$ to $AI$, and $T_2$ be the foot from $E$ to $AI$. Since $D$ is on line $AI$, $AE=AF$, and so $T_2$ is also the foot from $F$ to $AI$. This means that line $EF$ is the perpendicular from $T_2$ to $AI$, so we need to show that $T_1=T_2$. We compute $$AT_2^2-DT_2^2=AE^2-ED^2=(AD\cos\alpha)^2-(AD\sin\alpha)^2=AD^2(\cos^2\alpha-\sin^2\alpha)=AD^2\cos A,$$ where $\alpha=\frac12\angle A$. Now, by the law of sines, \begin{align*} \frac{AT_1^2-DT_1^2}{AD^2} &=\frac{AH^2-HD^2}{AD^2}\\ &=\frac{\sin^2(\angle ADH)-\sin^2(\angle HAD)}{\sin^2(\angle AHD)}\\ &=\frac{\sin(\angle ADH-\angle HAD)\sin(\angle ADH+\angle HAD)}{\sin^2(\angle AHD)}\\ &=\frac{\sin(\angle ADH-\angle HAD)}{\sin\angle AHD}, \end{align*} where we have used that $\sin^2x-\sin^2y=\sin(x-y)\sin(x+y)$. Now, let $A'$ be the point diametrically opposite $A$ on the circumcircle of $ABC$. Since $\angle BAH=\angle A'AC$, $AI$ is the bisector of $\angle HAA'$, and so $$\angle HA'A=\angle DA'A=(180^\circ-\angle ADA')-\angle DAA'=\angle ADH-\angle HAD,$$ and so the law of sines gives that $$\frac{AT_1^2-DT_1^2}{AD^2}=\frac{\sin(\angle ADH-\angle HAD)}{\sin\angle AHD}=\frac{\sin\angle HA'A}{\sin\angle AHA'}=\frac{AH}{AA'}.$$ Now, note that the circumcircles of $AHB$ and $ABC$ have the same radius, as they are reflections over $AB$. If this radius is $R$, this means that $AH=2R\sin\angle ABH=2R\cos A$. This means that $AH/AA'=\cos A$, as desired.

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  • $\begingroup$ This is still trig-bash. $\endgroup$
    – ZNatox
    May 13 at 11:39
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Here's a trig-based (but not exactly -bashed) solution, which OP is free not to accept. Perhaps someone will find clues to a fully-synthetic approach within.


First, a relation based on orthocenter $H$ and midpoint $M$ of $\overline{BC}$. Let $N$ and $H'$ be the feet of the perpendiculars from $M$ to $\overline{AB}$ and from $H$ to $\overline{MN}$.

enter image description here

Define $p:=|HH'|=|NB|$ (the equality follows from $M$ being a midpoint). Then simple right triangle trig tells us $$\left.\begin{align} |MN|&=\phantom{2}p\tan B \\ |H'N|&=2p\cot A \end{align}\right\} \;\to\; \tan\angle MHH' = \frac{|MN|-|H'N|}{|HH'|}=\tan B-2\cot A \tag1$$

Now, in the exercise proper described by OP, since $AI$ is the bisector of $\angle A$, the target result is equivalent to having $\overline{HE}\perp\overline{AD}$. I'll work in reverse, assuming this perpendicularity and proving that $D$ is collinear with $H$ and $M$.

So, define $E$ such that $\overline{HE}\perp\overline{AD}$, and let the perpendicular at $E$ meet the angle bisector at $D$. Let $K$ be the foot of the perpendicular from $H$ to $\overline{AB}$.

enter image description here

Defining $q:=|KE|$ and $A_2:=A/2$, we have $$|HK|=q\cot A_2 \tag2$$

Let $L$ be the point where bisector $\overline{AD}$ meets $\overline{HK}$. Observe that $L$ must be the orthocenter of $\triangle AHE$, as the intersection of the altitudes from $A$ and $H$; consequently, (the extension of) $\overline{LE}$ must be the third altitude, perpendicular to $\overline{AH}$ and therefore parallel to $\overline{BC}$. We conclude that $\angle LEK\cong\angle B$. We see then that $\overline{DE}$ decomposes into easily-computed lengths, and we have $$\begin{align} |DE|= q\tan B+q\tan A_2 \quad\to\quad \tan\angle HDD' &= \frac{|DE|-|HK|}{|KE|} \\[6pt] &=\tan B+\tan A_2-\cot A_2 \end{align} \tag3$$ Invoking one (and only one!) slightly-bashy trig identity $$2\cot2\theta = \cot\theta - \tan\theta \tag4$$ assures us that $\angle MHH'\cong\angle HDD'$ via equality of their tangents, from which we deduce the collinearity of $D$, $H$, $M$. $\square$

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  • $\begingroup$ Your solution is great! Thanks for sharing! $\endgroup$
    – ZNatox
    May 16 at 17:17

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