In a forum the following theorem was proposed, but I couldn't find a synthetic solution.

Given a quadrilateral $ABCD$, let $P$, $Q$, $R$ and $S$ be resp. the orthocenters of triangles $ABC$, $BCD$, $CDA$ and $DAB$. Prove that quadrilaterals $ABCD$ and $PQRS$ have the same area.

I found online a proof for the case $ABCD$ is cyclic, but have no clue for the general case.


Note that if $ABCD$ is cyclic a much stronger result holds: $ABCD$ and $PQRS$ are equal. This is problem 2.35 in Mathematical Olympiad Treasures by Andreescu and Enescu. They prove the theorem using vectors and derive it from a lemma which may be useful: if $H$ is the orthocenter of a triangle $ABC$ and $O$ its circumcenter, then one has: $\vec{OA}+\vec{OB}+\vec{OC}=\vec{OH}$.


Finding an expression for the coordinates of the orthocenter, as a function of the coordinates of the vertices, is not difficult. For instance, we get for $P$: $$ x_P=\frac{\left(y_A-y_B\right) \left(y_C-y_A\right) \left(y_B-y_C\right)-x_A x_B \left(y_A-y_B\right)-x_A x_C \left(y_C-y_A\right)-x_B x_C \left(y_B-y_C\right)}{x_C \left(y_A-y_B\right)+x_A \left(y_B-y_C\right)+x_B \left(y_C-y_A\right)},\\ \ \\ y_P=\frac{\left(x_A-x_B\right) \left(x_C-x_A\right) \left(x_B-x_C\right)-y_A y_B \left(x_A-x_B\right)-y_A y_C \left(x_C-x_A\right)-y_B y_C \left(x_B-x_C\right)}{y_A \left(x_B-x_C\right)+y_B \left(x_C-x_A\right)+y_C \left(x_A-x_B\right)}. $$ The signed area $S_{ABC}$ of triangle $ABC$ can be conveniently expressed as: $$ 2S_{ABC}=A\times B+B\times C+C\times A= x_A y_B-x_B y_A +x_B y_C-x_C y_B +x_C y_A -x_A y_C, $$ where I used the shortcut $A\times B=\det(A,B)$. Of course, the sign of $S_{ABC}$ depends on the order of the vertices. Also notice that denominators in the above formulas for $x_P$ and $y_P$ are just $-2S_{ABC}$.

We can then define a signed area for quadrilateral $ABCD$: $$ 2S_{ABCD}=S_{ABC}+S_{BCD}+S_{CDA}+S_{DAB}. $$ I used Mathematica to find general expressions for the signed areas of $ABCD$ and $PQRS$ as a function of the coordinates of $A$, $B$, $C$, $D$, and checked they are indeed equal. This counts as a proof, but the resulting expressions are too long and involved. Moreover, that gives no insight for the reason of that equality.

In doing the above, I also realized that a stronger equality holds, of which I hadn't been aware before. It turns out that

The area of triangle $PQR$ is the same as the area of triangle $DAB$ (that is of the triangle having $S$ as orthocenter), and so on.

Of course this implies that $ABCD$ and $PQRS$ have the same area. Hopefully, this stronger result could be easier to prove without using algebra and coordinates.

  • $\begingroup$ Tinkering with geogebra, I have noted these facts (I have proved none of them): 1) If $ABCD$ is a parallelogram, then so is $PQRS$. The same if $ABCD$ is convex, concave, rhombus, rectangle, trapezium, isosceles trapezium, kite. Well, the case of the rectangle is rather obvious. $\endgroup$
    – ajotatxe
    Aug 15 '17 at 14:37
  • $\begingroup$ I wonder if area calculation using vertex coordinates could do the trick. A fairly simple formula for the coordinates of the orthocenters can be found here. $\endgroup$
    – Jens
    Aug 15 '17 at 18:05
  • $\begingroup$ I used Mathematica to prove the theorem, but expressions are very long. However, I also found another area equality, which may be of help. See my edited question. $\endgroup$ Aug 16 '17 at 16:36
  • $\begingroup$ Glad to see the coordinate approach worked out. Does this approach also handle concave quadrilaterals? $\endgroup$
    – Jens
    Aug 16 '17 at 17:43
  • 1
    $\begingroup$ I, too, found a stronger result. I'll write $A^\prime$ for the orthocenter of $\triangle BCD$, etc., and let $K$ and $K^\prime$ be the intersections of the diagonals of the original and derived quadrilaterals, respectively. Then, $|\triangle AKB| = |\triangle A^\prime K^\prime B^\prime|$, etc. The coordinate approach to showing this isn't too horrible (with Mathematica's help) if one takes $K$ at the origin and $A:=(a,0)$, $B := (b \cos\theta, b\sin\theta)$, $C := (c,0)$, $D := (d\cos\theta, d\sin\theta)$. $\endgroup$
    – Blue
    Aug 17 '17 at 3:19

enter image description here To prevent this good question, which has been widely viewed but not yet answered, from expiring for lack of oxygen, I'll submit this imperfect proof. Perhaps someone can help me fill the gap, or see a way around it. Approaching the problem in its edited form, given quadrilateral $ABCD$, with diagonals $AC, BD$, and orthocenters $E,F,G$ of triangles $ABC, BCD$, and $CDA$, respectively, to prove without coordinate geometry that$$\triangle ABD=\triangle EFG$$Through $D$ and $E$ draw $DS$ and $ET$ parallel to $AB$ and $GF$, respectively, intersecting in $M$ (if they are parallel, $ABCD$ is a trapezoid and the proof is easier). Join $MA,MB,MG,MF$, and through $M$ draw $PMN$ perpendicular to $AB$, and $QMR$ perpendicular to $GF$, and join $QT$ and $PS$. Since $\angle SMP$ and $\angle TMQ$ are right angles, and $\angle PSM=\angle QTM$ (assuming that $S,T,U,V$ are concyclic), which seems to be the case but I lack proof), then triangles $MPS$ and $MQT$ are similar, making$$\frac{MT}{MS}=\frac{MQ}{MP}=\frac{MN}{MR}$$But since $AG$ and $BF$ are both perpendicular to $DC$, they are parallel, making$$\frac{MT}{MS}=\frac{GF}{AB}$$Therefore$$\frac{MN}{MR}=\frac{GF}{AB}$$and triangles $ABM$ and $MFG$, since their bases are in inverse ratio to their altitudes, are equal. But $$\triangle ABM=\triangle ABD$$since they are on the same base and in the same parallels. For the same reason$$\triangle MFG=\triangle EFG$$Therefore$$\triangle ABD=\triangle EFG$$Note: It might seem more direct to join $SB$ and $TF$, making $\triangle ABS=\triangle ABD$ and $\triangle GFT=\triangle GFE$, and then show that triangles $ABS$ and $GFT$, between parallels $AG$ and $BF$, are on bases $AS=GT$. But however much Geogebra indicates this equality of bases, I've yet to find proof of it.

  • $\begingroup$ Thank you for this proof. I had given up on this problem, but now I think I can reconsider it. $\endgroup$ Sep 11 '17 at 19:57

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