# Prove that points $E, H,$ and $F$ are collinear

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.

• 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 May 13 at 20:51
• You are right! I somehow convinced myself yesterday that it should follow up from it, I'll edit it out! May 13 at 21:08

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!

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.

• This is still trig-bash. May 13 at 11:39

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 $$\overline{MN}$$.

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 $$\overline{AB}$$.

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 $$D$$, $$H$$, $$M$$. $$\square$$

• Your solution is great! Thanks for sharing! May 16 at 17:17