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

$BB'$ and $CC'$ are heights of a given $\triangle ABC$ ($AB\ne AC$). $M$ is mid-point of $BC$ and $H$ is orthocenter of $\triangle ABC$ and $D$ is intersection of lines $B'C'$ and $BC$. Prove $DH \perp AM$. My idea is to prove that $AC'EHB'$ (or $C'BME$ or $B'EMC$) is cyclic. I tried lots of things, but I think the main thing I lacked in all my deductions if usage of $M$ being a mid-point.

Other idea is to prove that $H$ is orthocenter of $\triangle AMD$, since we already have $AH\perp MD$, it's enough to prove $MH\perp AD$. If $MH\cap AD=\{G\}$, then we should prove that $G$ lies on circle around $AC'(E)HB'$. So another idea would be to prove that some of triangles "composed" of $A,C',E,H,B',G$ with at least once vertices being $E$ or $G$ has circumradius $AH/2$. Some trigonometry?

Another idea would be to mark $E'$ as foot of height from $D$ to $AM$ and then use Menelaus theorem to prove that $A-H-E'$, that is, $E\equiv E'$.

EDIT. The only place where I think I can use the fact that $M$ is midpoint of $BC$ is that $BM=C'M=B'M=CM$, so there are some equal angles. It should help us prove $E$ lies on circumcircle of $\triangle C'BM$, that is, $C'BME$ is cyclic, because we have $\angle MC'B=MBC'$. It's also interesting to notice that once we prove $E$ lies on mentioned circle, we'd conclude that $DE$ is bisector of $\angle C'EB$, which might be a clue but I still can't figure anything out.

EDIT 2. I also thought about using the fact that $\triangle CMA$ and $\triangle BMA$ have same areas. Then we have $AC\sin\alpha_1=BC\sin\alpha_2$ where $\alpha_1=\angle CAM=\angle CAE$ and $\alpha_2=\angle BAM=\angle BAE$, so we might be able to use that somehow for the idea about circumradius being $AH/2$, because we can use sine laws on triangles $\triangle AC'E$ and $AEB'$ (we need to prove $\angle C'EA=\angle B=\beta$ and/or $\angle B'EA=\angle C=\gamma$), from which we might reduce this to proof about similarity of some triangles. Seems like a nice idea that could work but I can't seem to get much from it for now.

Another thing, about the idea of proving that $H$ is orthocenter of $\triangle AMD$. I'm not sure is it better to mark $MH\cap AD=\{G\}$ and prove $MH\perp AD$, or other way around? Or maybe to mark intersection of $AD$ and circumcirle of $AC'HB'$ and $\triangle ABC$ (maybe this is important?) as $G$ and then prove $H\in GM$?

Just adding ideas, since I can't seem to get anywhere, almost like we lack just a little bit something and all of the ideas might actually work very similarly.

Also, since people wondered if it's Olympiad problem or homework problem: it's from Serbian mathematical magazine for high schoolers, it's in the section where one of the hardest problems are. But I still find it harder than the usual ones from geometry.

Please no analytic geometry or complex numbers, and try avoiding vectors unless used only for less significant lemmas.

• Looks like an olympiad problem. – Joe Z. Mar 14 '13 at 20:15
• Is this homework? – PyRulez Mar 14 '13 at 21:18
• @PyRulez, we should celebrate if problems of this difficulty are being used as homework. – zyx Mar 14 '13 at 21:26
• @LazarLjubenović [spam]500 poena, jesi lud :) Pozdrav za tebe i nadam se da ces naci resenje (u sta ne sumnjam, ima ovde dobrih matematicara), jos kad vide ovolike poene, ih :)[/spam] – Cortizol Mar 14 '13 at 22:56
• ^ I read the above by feeding it into Google Translate. "Jesi lud" indeed. – Joe Z. Mar 15 '13 at 13:59

## 5 Answers

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.

• Thanks a lot. This is the kind of solution I needed. Very nice usage of trigonometry, I couldn't get anywhere near it although it seems so simple now. Nice general idea as well. – Lazar Ljubenović Mar 19 '13 at 19:23
• @LazarLjubenović: Thanks a lot to you! :) I suspect that some of the insights in other answers could streamline my argument a bit, but at least this approach shows what you can do with a little angle-chasing. – Blue Mar 19 '13 at 22:59
• Please take a look on the following question: math.stackexchange.com/questions/2321816/… – Blind Jun 15 '17 at 21:14

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.

• (+1) nice approach. Can you elaborate more on the involution? It seems to me your involution is one which first transform $A$ to $\infty$ and $g$ to a diameter $g'$ of the circle, you then reflect everything w.r.t $g'$. Finally, you transform $\infty$ and $g'$ back to $A$ and $g$ respectively. Is that correct interpretation? – achille hui Mar 15 '13 at 16:04
• Thanks, this is very clever. – achille hui Mar 15 '13 at 16:31
• You could view it as an inversion with radius of inversion equal to the power of $A$ w.r.t. the circle. – Ivan Loh Mar 15 '13 at 16:44
• I take it that the question asker was looking for an elementary geometry solution; do projective planes count? – Joe Z. Mar 15 '13 at 18:04
• Very nice. Once the diagram is drawn in that way, the solution is as simple as saying "the polar of a point with respect to a circle, is perpendicular to the line from that point to the center". – zyx Mar 16 '13 at 0:07

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.

this problem is really orthogonal circles, let us from pure geometry view to solve it.

first to find two orthogonal circles from triangle ABC. see picture below: look BC'B'C: ${CC'} \perp {AB}$,$BB' \perp AC$,so BC'B'C is cyclic and M is the center.

look at AC'HB',they are also cyclic. the center is the N of midpoint of AH.

BTW,A'C'NBM is cyclic which is the 9 points circlehttp://en.wikipedia.org/wiki/Nine_points_circle and $C'N \perp C'M$ as NM is the diameter.

so circle NA and circle MC are orthogonal circles.

2nd: we will proof if DM is the diameter to make a circle, then this circle and circle NA are also orthogonal circles. I use another picture to proof it. suppose circle A and B are orthogonal circles, C and D are the two common points, then $AD \perp BD$,$AC \perp BC$. line AB,CD, cross at E. then $AB \perp CD$. let $AD=p,BD=q$, then $AB= \sqrt{p^2+q^2}=d$

take F on CD,$FE=x, AE=m= \dfrac{p^2}{d}, BE=n=\dfrac{q^2}{d}$, line AF, G is midpoint of AF, $GF=r=\dfrac{AF}{2}=\dfrac{\sqrt{x^2+m^2}}{2}$,make $GH \perp AB,$ foot H, then $GH= \dfrac{FE}{2}=\dfrac{x}{2},HE=\dfrac{m}{2},GB^2=GH^2+HB^2=(\dfrac{x}{2})^2+(\dfrac{m}{2}+n)^2=\dfrac{x^2}{4}+\dfrac{m^2}{4}+mn+n^2)=r^2+n(m+n)=r^2+nd=r^2+q^2$.

so circle GF and circle B are orthogonal circles. look at diameter JK of circle A, please note A is midpoint of JK.BTW, circle G(red) must through E.

3rd: orthogonal circles have some interesting fact which already proof, see another picture: circle P and Q are orthogonal, then take A in circle P, make AB is diameter,E and F are common points of circle P and Q, AE meet circle Q at C, then $CQ \perp AB$,and CQ cross circle Q at D, BC must through E, D must the orthocenter of ABC.

I left detail proof as exercise , you can find details from "Notes On the Orthcentric Tetrahedron" paper on web ,which is recommended in SE.

last step: see final picture: K is midpoint of DM, make circle KM, then circle KM and circle NA are orthogonal circles. look at triangle ADM, （you may need to proof A must be the point which equal to 3rd step point C,it is not difficult when you finish 3rd step.)$AN \perp DM$, cross circle NA　at H, so H must be the orthocenter　of ADM. that is done.

I add detail proof for the last step:

circle KD have common points E and F. line KE,NE, since two circles are orthogonal circles which ii proof in step2, so $KE \perp NE$,line ME ,suppose cross at circle NA at A2, line A2N, $\angle NA2E=\angle NEA2,\angle KEM=\angle KME, \angle KEM+\angle NEA2=\dfrac{\pi}{2},$ so $\angle KME+\angle NE2A=\dfrac{\pi}{2}$,that is $A2N \perp DM$, since both AN and A2N have common point N, A2 must be A. with same reason, DF extend circle NA, it must cross at A. if we connect DE,HE,since DM is diameter, so $DE\perp AM$, since AH is diameter, so $HE \perp AM$, both have common point E, that means H must on DE. and H must orthcenter of triangle ADM. that is finished.

• Those are very nice pictures. What did you use to make them? – davidlowryduda Mar 16 '13 at 14:35
• it is geopad which popular for geometry solution. – chenbai Mar 17 '13 at 4:25
• Please take a look on the following question: math.stackexchange.com/questions/2321816/… – Blind Jun 15 '17 at 21:14

Note that the partial solution here is not actually an answer (because I didn't get to it) – I have left it here for the benefit of people who would like to see a thought process into the problem.

Note that $BC'B'C$ is actually cyclic, because $BC'C$ and $BB'C$ are both right triangles. Then $M$ is the center of the circle that circumscribes $BC'B'C$, because it is the midpoint of diameter $BC$.

Playing around with GeoGebra showed me that the tangent points from $A$ to the circumcircle of $BCB'C'$ (the tangent points of the two tangent lines from $A$ to the circle) lie on $DH$. Let these two points be $T_1$ and $T_2$. If we use this fact, we see that $AT_1 = AT_2$ and $MT_1 = MT_2$, so $AT_1MT_2$ is a kite, and thus $AM$ and $T_1T_2$ are perpendicular; therefore $AM$ and $DH$ are also perpendicular (because $DH$ and $T_1T_2$ are the same line).

I think this is the most natural use of $M$ being the midpoint of $BC$ that we can find. If we can somehow prove that fact (that the tangent points from $A$ to the circumcircle of $BCB'C'$ lie on $DH$), then we are done.

Notes for trying:

• By the power-of-a-point theorems, $(AT_1)^2 = AB' \cdot AC = AC' \cdot AB = (AT_2)^2$.
• Christian Blatter seems to have proven an equivalent fact using projective geometry in his answer. – Joe Z. Mar 15 '13 at 20:28
• That's a very big bounty o_O – Joe Z. Mar 23 '13 at 1:01