In a convex quadrilateral $ ABCD$ the points $ P$ and $ Q$ are chosen on the sides $ BC$ and $ CD$ respectively so that $ \angle{BAP}=\angle{DAQ}$. Prove that the line, passing through the orthocenters of triangles $ ABP$ and $ ADQ$, is perpendicular to $ AC$ if and only if the triangles $ ABP$ and $ ADQ$ have the same areas.
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Tags: geometry, circumcircle
RSM
07.04.2012 23:45
Under inversion wrt $ A $, suppose, any point $ L $ maps to $ L' $. Suppose, $ H_1,H_2 $ are the orthocenters of $ \Delta ADQ $ and $ \Delta ABP $. $ AH_1' $ passes through the circumcenter of $ \Delta AD'Q' $(call it $ O_1 $) and circumcenter of $ \Delta AB'P' $ be $ O_2 $. $ AC\perp H_1H_2 $ implies that the radical axis of $ \Delta AD'Q' $ and $ \Delta AB'P' $ passes through the circumcenter of $ \Delta AH_1'H_2' $. So $ H_1'H_2'O_1O_2 $ is cyclic. So $ AH_1'.AO_1=AH_2'.AO_2\implies AD.AQ=AB.AP\implies [ADQ]=[ABP] $. So done.
TheBernuli
03.03.2014 21:16
$AC\perp H_1H_2$ implies that the radical axis of $\Delta AD'Q'$ and $\Delta AB'P'$ passes through the circumcenter of $\Delta AH_1'H_2'$. I can't understand this..
EugenyVassilievitch
24.01.2017 21:02
Actually, it is enough to show that $AH_{1}^2 + CH_{2}^2 = AH_{2}^2 + CH_{1}^2$, to prove that the shown lines are perpendicular to each other. After that we just need to prove that this equality is equivalent to $AD*AQ = AB*AP$. It can be done by some simple calculations.