Prove that line through intersection of line through foots of heights and opposite line and orthocenter is perpendicular to median.

Solution 1:

Assume the given triangle $ABC$ is acute. Draw the circle with center $M$ through $B$, $C$, $B'$, $C'$; then draw tangents from $A$ (which lies outside) to the circle. Draw a red line $g$ through the two tangent points; this line intersects the line $B\vee C$ in a point $D$.

Our drawing plane can be embedded in a projective plane $P$ by adding elements at infinity (we won't need them). At any rate there is a perspective involution $\iota:\ P\to P$ with center $A$ which keeps $g$ pointwise fixed, and which has the following additional property: For any line $\ell$ through $A$ that intersects the circle it interchanges the two intersection points. (Projectively, this $\iota$ is conjugate to an ordinary reflection $\iota'$ in the $x$-axis $g'$. The map $\iota'$ has its center $A'$ at $y=\pm\infty$, keeps $g'$ pointwise fixed and interchanges points on the unit circle that lie on the same vertical.)

In particular $\iota(B)=C'$, $\iota(C')=B$, $\iota(C)=B'$, $\iota(B')=C$. Therefore $\iota(C'\vee B')=B\vee C$. Therefore $C'\vee B'$ and $B\vee C$ intersect on $g$, which implies that our point $D$ is the same as the point $D$ in the question. In the same way it follows that $B\vee B'$ and $C\vee C'$ intersect on $g$, which implies that the point $H$ lies on $g$ as well.

Since the line $A\vee M$ is orthogonal to $g$, it is orthogonal to $D\vee H$, as stated.

Solution 2:

Let $A^\prime$ be the foot of the altitude (through $H$) from $A$.

Approach: We'll use proportionality to show that $\triangle MA^\prime A\sim\triangle HA^\prime D$, whereupon (via Angle-Angle) $\triangle HA^\prime D \sim \triangle EHA$, so that $\angle HEA \cong \angle HA^\prime D$ (a right angle).

To unify the trig, let $k$ be the circumdiameter of $\triangle ABC$. Thus,

$$|BC| = k \sin A = k\sin(B+C) \qquad |CA| = k \sin B \qquad |AB| = \sin C$$

(Throughout, "$\angle A$", "$\angle B$", and "$\angle C$" refer to the angles in the corners of $\triangle ABC$.) We'll show $\triangle MA^\prime A \sim \triangle H A^\prime D$ by showing

$$ \frac{|MA^\prime|}{|AA^\prime|} = \frac{|HA^\prime|}{|DA^\prime|}$$

Let's take each element of the proportion in turn. Two are easy:

$\begin{align} |MA^\prime| &= |MC| - |A^\prime C| \\ &=\frac{1}{2}|BC| - |CA|\cos C \\ &=\frac{k}{2}\left( \sin(B+C) - 2 \sin B \cos C \right)\\ &=\frac{k}{2}\sin(C-B) \\[8pt] |AA^\prime| &= |AB| \sin B \\ &= k \sin B \sin C \end{align}$

For $|HA^\prime|$, first note that $\triangle A^\prime HB \sim \triangle B^\prime CB$, as these are two right triangles that share an acute angle (at $B$). Therefore, $\angle A^\prime H B \cong \angle C$, and we have

$$|HA^\prime| = |A^\prime B| \cot C = |AB| \cos B \cot C = k \sin C \cos B \cot C = k \cos B \cos C$$

Segment $|DA^\prime|$ is the interesting one. First, we write $$|DA^\prime| = |CA^\prime| + |DC| = |AC|\cos C + |DC| = k \sin B \cos C + |DC|$$

Then, by the Law of Sines, $$\frac{|DC|}{|CB^\prime|} = \frac{\sin \angle DB^\prime C}{\sin \angle B^\prime DC} \implies |DC| = |BC| \cos C \cdot \frac{\sin \angle DB^\prime C}{\sin \angle B^\prime DC} = k \sin A \cos C \cdot \frac{\sin \angle DB^\prime C}{\sin \angle B^\prime DC}$$

To get a handle on those final angles, it helps to have a picture.

We'll use the fact that $AB^\prime H C^\prime$ lie on a circle (with diameter $AH$, which is subtended by right angles at $B^\prime$ and $C^\prime$). Note that $\angle BAA^\prime$ is the complement of $\angle B$ in $\triangle AA^\prime B$. In the circle, $\angle BAA^\prime$ and $C^\prime B^\prime H$ subtend the same chord, $C^\prime H$, and are therefore congruent. Since $\angle C^\prime B^\prime H$ and $\angle A B^\prime C^\prime$ form a right angle at $B^\prime$, we have that the latter must be congruent to $\angle B$; further $\angle DB^\prime C \cong \angle A B^\prime C^\prime$ as vertical angles, so that $\angle DB^\prime C \cong \angle B$. Moreover, $\angle DCB^\prime$ is the supplement of $\angle C$, and we have $\angle CDB^\prime = \angle C - \angle B$, whence

$$|DC| = \frac{k \sin(B+C) \sin B \cos C}{\sin(C-B)}$$

and

$$\begin{align} |DA^\prime| &= k \sin B \cos C + \frac{k\sin(B+C)\sin B\cos C}{\sin(C-B)} \\ &= \frac{k \sin B \cos C \; \left(\sin(C-B)+\sin(B+C)\right)}{\sin(C-B)} \\ &= \frac{2 k \cos B \sin B \cos C \sin C}{\sin(C-B)} \end{align}$$

After all of this, we see that, in fact

$$\frac{|HA^\prime|}{|DA^\prime|} = k \cos B \cos C \cdot \frac{\sin(C-B)}{2k\cos B\sin B\cos C\sin C} = \frac{\sin(C-B)}{2\sin B\sin C} = \frac{|MA^\prime|}{|AA^\prime|}$$

completing the proof.

Solution 3:

This is a special case of the following well known result:

Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$. Let $BD$ intersect $AC$ at $H$, $AB$ intersect $CD$ at $E$, $BC$ intersect $AD$ at $F$. Then $H$ is the orthocenter of triangle $EOF$.

Proof: (Harmonic Conjugates and Poles and Polar): It suffices to prove that $OE \perp HF$, as by symmetry we will have $OF \perp HE$ as well, so that $H$ is indeed the orthocenter of $EOF$. Let $FH$ intersect $CD$ at $X$, $AB$ at $Y$. Then $D, X, C, E$ are harmonic, and so are $A, Y, B, E$. Now this implies that $XY$ is the polar of $E$ with respect to the circumcircle, so we indeed have $OE \perp HF$, and we are done.

Edit: To see how this finishes your problem, apply it to cyclic quadrilateral $BCB'C'$ with circumcenter $M$.

In the above proof, I implicitly used the following properties about harmonic conjugates and poles and polar. In case you are unfamiliar with this topic, I have provided the proofs for them as well. Apart from the above links I provided, you can try googling "Harmonic conjugates" and "Poles and Polar" to find more information.

Property 1: In the diagram below, $A, B$ and $C, D$ are harmonic conjugates.

Proof: By Ceva's theorem on triangle $LAB$ we have $\frac{AD}{DB}=\frac{NL}{BN}\frac{MA}{LM}$. By Menelaus's theorem on triangle $LAB$ and line $MNC$ we have $\frac{AC}{CB}=\frac{NL}{BN}\frac{MA}{LM}$. Thus $\frac{AD}{DB}=\frac{AC}{CB}$ so $A, B$ and $C, D$ are harmonic conjugates. One could also say that $A, D, B, C$ form a harmonic range.

Property 2: If line $l$ is the polar of point $P$ with respect to a reference circle with center $O$, then $OP \perp l$.

Proof: Obvious by definition of poles and polar.

Property 3: Consider a point $P$ outside a reference circle with center $O$, and let $PA, PB$ be tangents to the reference circle. Then $AB$ is the polar of $P$ with respect to the reference circle.

Proof: Let $OP$ intersect $AB$ at $C$. It is clear that $OP \perp AB$. Now triangle $OPA$ is similar to triangle $OAC$, so $\frac{OA}{OP}=\frac{OC}{OA}$, so $OP \cdot OC=OA^2$, so $C$ is the image of $P$ upon inversion about the reference circle, so $AB$ is the polar of $P$ with respect to the reference circle.

Property 4: Consider a reference circle with center $O$, and a point $P$. Let the polar of $P$ with respect to the reference circle be $l$. Draw a line through $P$ intersecting the reference circle at $X, Y$ and intersecting $l$ at $Z$. Then $P, Z$ and $X, Y$ are harmonic conjugates.

Proof: Draw tangents $PA, PB$ from $P$ to the reference circle, then $AB$ is the line $l$. Let $AB$ intersect $OP$ at $C$, then $OC \cdot OP=OX^2$, so triangle $XOP$ is similar to triangle $COX$. Now, $\frac{XP}{OX}=\frac{CX}{OC}$. Similarly, $\frac{YP}{OY}=\frac{CY}{OC}$. Dividing gives $\frac{XP}{YP}=\frac{CX}{CY}$. We can similarly show $\frac{XZ}{YZ}=\frac{CX}{CY}$, so indeed $P, Z$ and $X, Y$ are harmonic conjugates.