I been doing some geometry of some problems from Olympiads. I found one that I really haven't been able to solve. It's been 3 days since I
found it. I have to admit that I haven't been all the time doing this problem, but I've spend around 3 and half hours doing this problem and I couldn't really crack the nut.
In an acute triangle $ABC$, an arbitrary point $P$ is chosen on the altitude $AH$. The points $E$ and $F$ are the midpoints of $AC$ and $AB$, respectively. The perpendiculars from $E$ on $CP$ and from $F$ on $BP$ intersect at the point $K$. Show that $KB = KC$.
Well, at first I tried Euclidean geometry and tried to use the midpoint theorem and some constructions and I even discovered something cool about the circle with radius $KC$, then I realized I could prove that $K$ lies on the perpendicular bisector of $BC$, use the Pythagorean theorem and I even noticed that the point $K$ was the intersection of 3 perpendicular lines, so I could hypthecally prove that $K$ was the circuncenter of some triangle involving these sides.
At the end , I did use Cartesian coordinates, but, there were so many equations and really, didn't get do much satisfaction.
Does someone has some good idea on how to solve it with Euclidean geometry?
I pretty much did all of that.
Best Answer
Let $E',F'$ the midpoint of $BP, CP$ and $K'$ the circumcenter of $BCP$. Then we can show $|EE'|=|FF'|=\frac{1}{2}|AP|$ and both lines are perpendicular to $BC$ using the Intercept theorem. Because $EK,FK$ are parallel to $E'K',F'K'$ we conclude that $KK'$ is perpendicular to $BC$ and therefore $K$ lies on the perpendicular bisector of $BC$.