Let $ABC$ be an acute triangle with $\angle C>\angle B$. Let $D$ be a point on $BC$ such that $\angle ADB$ is obtuse, and let $H$ be the orthocentre of triangle $ABD$. Suppose that $F$ is a point inside triangle $ABC$ that is on the circumcircle of triangle $ABD$. Prove that $F$ is the orthocenter of triangle $ABC$ if and only if $HD||CF$ and $H$ is on the circumcircle of triangle $ABC$.
Problem
Source: Chinese MO 1999
Tags: geometry, circumcircle, geometry proposed
hatchguy
19.03.2012 23:00
Suppose that $F$ is the orthocenter of $ABC$. Clearly, since $HD \perp AB$ and $CF \perp AB$ then $HD \parallel CF$. Also we have $\angle BDA= \angle BFA = 180 - \angle C$ and therefore $\angle BHA = \angle C \rightarrow H$ is on the circumcircle of $ABC$. Suppose $H$ is on the circumcircle and $CF \parallel HD$. Clearly we have $CF \perp AB$. Also, since $H$ is on the circumcircle $\angle BHA = \angle C \rightarrow \angle BDA = 180 - \angle C \rightarrow \angle BFA = 180 - \angle C$ and therefore $F$ is the orthocenter.
Pedram-Safaei
25.03.2012 14:53
it is very simple and interesting,and solved easily by angle chasing,...
sayantanchakraborty
25.06.2014 09:42
Its all about going through the proofs carefully and not mixing up what is given and what is required to prove.Let $F$ be the orthocenter of $ABC$.Then by hypothesis $AF \perp BC$ and $AH \perp BC \Rightarrow A,F,H$ are collinear.Again $CF \perp AB$ and $HD \perp AB \Rightarrow CF \parallel HD$.Now $\angle{FDC}=90^{\circ}-\angle{HFD}=90-B=\angle{CFD}$.Enjoy! Now the other part:Assume that $H$ lies on the circumcircle of $\triangle{ABC}$.Let $X=AH \cap BC$ and $F'=\odot{ADB} \cap AH$.By hypothesis $AX \perp BC$.Also easy to note that $\angle{XHD}=B$ and $\angle{XF'D}=\angle{ABD}=B$.So $F' \equiv F$.Enjoy!