Let $ABC$ be a triangle and $H$ its orthocenter. Let $D$ be a point lying on the segment $AC$ and let $E$ be the point on the line $BC$ such that $BC\perp DE$. Prove that $EH\perp BD$ if and only if $BD$ bisects $AE$.
Problem
Source: 2019 Baltic Way P12
Tags: geometry
MarkBcc168
18.11.2019 16:34
Let $P=BD\cap AH$. Since $DE\parallel AH$, we get $BD$ bisects $AE$ if and only if $AD\parallel EP$. But this is equivalent to $H$ being the orthocenter of $\triangle BEP$ which is equivalent to $EH\perp BD$.
timon92
20.11.2019 21:01
This problem was proposed by Burii.
Gonzo17
21.11.2019 00:19
There's also this one-liner (if one knows his lemmas): From the Gauss-Bodenmiller Theorem on the quadrilateral ABED BD bisecting AE is equivalent to BD being perpendicular to the line formed by orthocenters of ABC, DEC, which is EH.
arzhang2001
10.08.2020 07:34
another solution : let $S$ be foot of altitude from $C$ to $AB$ . And let $HE$ Intersect $BD$ At $T$ if $HE \perp BD$ Then$STDA$ cyclic. and we have: $BS.BA=BT.BD=BE^2$ now we get if $HE \perp BD$ should we have this condition. Now suppose that $BD$ bisect $AE$ . we have : $\frac {AD}{DC} = \frac {BE}{BC}$ So if we draw altitude $AK$ TO $BC$ We should have: $\frac {KE}{EC} = \frac {BE}{BC}$ hence :$\frac {BE}{BC} = \frac {BE-BK}{BC-BE} \implies BE^2=BK.BC=BS.BA$ and we are done . $\blacksquare$
Tsikaloudakis
17.04.2023 10:31
στις υποψιν;