Let $ABCD$ be a convex quadrilateral with non perpendicular diagonals and with the sides $AB$ and $CD$ non parallel . Denote by $O$ the intersection of the diagonals , $H_1$ the orthocenter of the triangle $AOB$ and $H_2$ the orthocenter of the triangle $COD$ . Also denote with $M$ the midpoint of the side $AB$ and with $N$ the midpoint of the side $CD$ . Prove that $H_1H_2$ and $MN$ are parallel if and only if $AC=BD$
Problem
Source: Romania JBMO TST 2015 Day 1 Problem 5
Tags: geometry, quadrilateral, orthocenter, parallel
TelvCohl
14.05.2015 15:41
My solution : Let $ X, Y $ be the midpoint of $ BC, DA $, respectively . Since $ H_1H_2 $ is the Steiner line of complete quadrilateral $ \{ AB, CD, AC, BD \} $ , so $ H_1H_2 $ is perpendicular to the Newton line $ XY $ of the complete quadrilateral $ \{ AB, CD, AC, BD \} $ . Since $ MXNY $ is a parallelogram ( $ MX=NY=\tfrac{1}{2} AC, MY=NX=\tfrac{1}{2} BD $ ) , so $ H_1H_2 \parallel MN \Longleftrightarrow MN \perp XY \Longleftrightarrow MXNY $ is a rhombus $ \Longleftrightarrow AC=BD $ . Q.E.D
kjytay
06.08.2015 04:36
Label the feet of the perpendiculars and extend them to meet at points $X$ and $Y$ as in the diagram:
Define 4 circles: $\Gamma_1 = \odot AA'GD, \Gamma_2 = \odot DCGF, \Gamma_3 = \odot EFCB, \Gamma_4 = \odot AEA'B$ (vaguely labelled in diagram as well).
Consider the radical axis for each pair of circles. Note that for any 3 circles, the 3 radical axes they generate must meet at a point. By going through all the pairs of circles, we can conclude that $XY$ is the radical axis of $(\Gamma_2, \Gamma_4)$. Since $M$ and $N$ are the centres of $\Gamma_2$ and $\Gamma_4$ respectively, we have $XY \perp MN$.
Next, since $AA'$ and $CF$ are both perpendicular to $BD$, we have $H_1X \parallel H_2Y$. Similarly, $H_1Y \parallel H_2X$ implying that $H_1XH_2Y$ is a parallelogram.
Finally, note that $A'F = AC \cos AOB$ and $EG = BD \cos AOB$.
Putting this all together:
$AC = BD \\
\Leftrightarrow A'F = EG \\
\Leftrightarrow Area(H_1XH_2Y) / H_1X = Area(H_1XH_2Y) / H_2X \\
\Leftrightarrow H_1X = H_2X \\
\Leftrightarrow H_1XH_2Y \text{is a rhombus} \\
\Leftrightarrow XY \perp H_1H_2 \\
\Leftrightarrow MN \parallel H_1H_2.$