In an acute-angled triangle $ABC$ with $AB < AC$, the circle $\omega$ touches $AB$ at $B$ and passes through $C$ intersecting $AC$ again at $D$. Prove that the orthocentre of triangle $ABD$ lies on $\omega$ if and only if it lies on the perpendicular bisector of $BC$.
Problem
Source: RMO Problem 3
Tags: geometry, perpendicular bisector, geometry unsolved
11.12.2013 08:05
Let $H$ be the orthocenter of $\triangle ABD$. $H$ lies on the perpendicular bisector of $BC\implies \angle ADB=\angle ABC\implies \angle ADH+\angle BDH=\angle ABH+\angle HBC\implies$ $\angle BDH=\angle HBC=\angle BCH$. $H$ lies on $\omega\implies \angle ABH=\angle BCH=\angle BDH=\angle BAH\implies AH=BH\implies$ $AD=BD\implies \angle CBH=\angle ADH=\angle BDH=\angle BCH\implies BH=CH$,as desired. $Q.E.D.$
20.11.2015 20:00
Hey Mathlinkers, Sorry for reviving this thread. I have an observation in the problem If $H$ lies on $perpendicular$ $bisector$ then it also lies on the $\omega$.. PROOF Consider the triangle $BDC$. Let the $orthocentre$ be $H$. Drop perpendiculars from $A,B$, to $BD,AD$ Also draw the perpendicular bisector......Now observe that $perpendiculars$ are being dropped from the intersection of the perpendicular bisector and $\omega$ to the sides $ad,BD,BC$(a point on the $circumcircle$) to the sides of the triangle $BDC$ so by the $simson$ $Line$ $theorem$ we have that the feet of perpendiculars from that point are collinear... Can this Help? Please don't downvote if it doesn't help
20.11.2015 20:34
A concrete solution using angle chasing.. Note that ∠$ADB =$ ∠$B$ and hence triangles $ADB$ and $ABC$ are similar. In particular, $ABD$ is an acute-angled triangle. Let H denote the $orthocenter$ of triangle $ABD$. Then ∠$BHD$ = $180$◦ − ∠$A$. Suppose that $H$ lies on Γ. Since $4AB < AC$ the point $D$ lies on the segment $AC$ and ∠$C = 180$◦ − ∠$BHD = $∠$A$. Therefore $BH$ is the perpendicular bisector of $AC$. Hence ∠$HBC$ = ∠$ABC$ = ∠$HCB$, so $H $lies on the perpendicular bisector of$ BC$. Conversely, suppose that $H $lies on the perpendicular bisector of $BC$. Then ∠$HCB$ = ∠$HBC$ = $90$− $\angle C$. Since ∠$ABD = $∠$C$ it follows that ∠$HDB = $$90$◦ − ∠$C$. Since ∠$HCB = $∠$HDB$ we have that $H$ lies on Γ ....... QED
17.09.2016 20:12
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05.10.2020 23:34
file:///C:/Users/arjun/OneDrive/0_Arjun/IMO/RMO%20past%20year%20papers/my%20solution,%20RMO%202013-3%20(paper%203).pdf Go to above link to check my solution to the problem. I used simson line theorem .
31.05.2024 17:20
i had done this problem after completing liog ch1 power of point in my mind there is power of point i had made construction and found radical axis but that is not useful after that i had made an simple observation