Let $ABC$ be an acute triangle. The circumference with diameter $AB$ intersects sides $AC$ and $BC$ at $E$ and $F$ respectively. The tangent lines to the circumference at the points $E$ and $F$ meet at $P$. Show that $P$ belongs to the altitude from $C$ of triangle $ABC$.
Problem
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Tags: geometry, circumcircle, trigonometry, perpendicular bisector
NT2048
13.08.2013 04:51
We will prove that $\angle PCF=90-\angle B$ by showing that $P$ is the circumcentre of $\triangle CEF$.
First of all, since diameters subtend right angles, $F$ and $E$ are the feet of the altitudes. We have $\angle BAF=90-\angle B$ and $\angle ABE=90-\angle A$. Now if $H$ is the orthocentre (intersection of $AF$ and $BE$), $\angle EHF=\angle AHB=\angle A+\angle B$.
By the alternate segment theorem, $\angle PFA=\angle B$ and $\angle PEB=\angle A$. In quadrilateral $PEXF$, we find that $\angle EPF=360-2(\angle A+\angle B)=2\angle C$. By equal tangents, $PF=PE$.
Consider the circle centred at $P$ with radius $PF$. Since $\angle EPF=2\angle ECF$ as we showed above, $C$ lies on this circle. This means that $PF=PC$.
Now just angle chase. We want to prove that $\angle PCB=90-\angle B$, but this is clearly true since $\angle PFC=90-\angle PFA=90-\angle B$.
Edit: Yeah, much of the stuff at the end is unnecessary. Just notice that HEFC is cyclic and due to right angles $HC$ is a diameter, so it passes through the centre, meaning that $P$ lies on the altitude.
sunken rock
13.08.2013 18:02
A shorter way to finish: since $BCEF$ is cyclic, $\Delta AEF\sim\Delta ACB$ and so the circumcenter of $\Delta AEF$ lies onto the altitude from $A$ of $\Delta ABC$. Best regards, sunken rock
ThirdTimeLucky
13.08.2013 18:31
An even shorter way is by poles and polars, for if $ EF \cap AB= G $, the polar of $ G $ passes through $ P $ and $ P $ belongs to altitude $ CH $ by brokards theorem, where $ H $ is the orthocenter.
mathdebam
13.08.2013 19:41
Longer way by trigo: Basically I first show that $P$ is circumcentre of $EFC$.To show it I use trigo.Applying sine rule to $EFP$ you get $PE=PF=\frac{a cosC}{2sinC}=RcosC$ where R is circumradius of $ABC$.Again circumradius of $EFC$ $r=\frac{EC}{2sinEFC}=\frac{BCcosC}{2sinA}$(why?)$=RcosC$.Thus $PE=PF=r$.Now the circumcentre of the triangle$EFC$ $O$ must lie on the perpendicular bisector of $EF$ and $OE=OF=r$. But there is only a unique such point $O$ and the fact that $P$ satisfies the condition shows that $P=O$i.e $P$ is the circumcentre of $EFC$.So$PE=PC$.But a mere angle chasing will yeild that $PEC=90-A$.So $PCE=90-A$.Thus $CP$ is the altitude or $P$ lies on the altitude from $C$. I think that GEOMETRY is the way of symbolizing the endless colourful beauties of nature through the unicolured pencils,pens and papers.
Tumon2001
18.03.2017 11:58
A bit different: Consider a $\Delta ABC$ with altitudes $CD$, $AF$ and $BE$. Let the tangent to $(AEFB)$ at $E$ meet $CD$ at $P$. Join $PF$. It would be enough to prove that $PF$ is tangent to $(AEFB)$ or, more specifically, $PE = PF$. We have $\angle PEF = \angle CAF = \angle CDF$. So, $PEDF$ is cyclic. Since, $CD$ bisects $\angle EDF$, the result follows.
babu2001
20.03.2017 09:43
Applying Pascal's Theorem on $BEECFF $, we get $EE\cap FF\equiv P $ lies on $AH $, where $BE\cap CF\equiv H $, so $H$ is the orthocentre $\implies P $ loes on the altitude through $A $.