2007 Turkey MO #1

In an acute triangle $ABC$, the circle with diameter $AC$ intersects $AB$ and $AC$ at $K$ and $L$ different from $A$ and $C$ respectively. The circumcircle of $ABC$ intersects the line $CK$ at the point $F$ different from $C$ and the line $AL$ at the point $D$ different from $A$. A point $E$ is choosen on the smaller arc of $AC$ of the circumcircle of $ABC$ . Let $N$ be the intersection of the lines $BE$ and $AC$ . If $AF^{2}+BD^{2}+CE^{2}=AE^{2}+CD^{2}+BF^{2}$ prove that $\angle KNB= \angle BNL$ .


Solution 1

Because $BC \perp AD$ we have that $BD^2+AC^2=DC^2+AB^2$ , and because $ FC \perp AB$ we have that $ FA^2+BC^2=FB^2+AC^2$. Combining the equalities above with the given one we see that $ CE^2+AB^2=AE^2+BC^2$, implying $ BE \perp AC$. Therefore, $ BN$ is the third height of $ ABC$.  It is well-known the altitudes of the triangle are the internal bisectors of its orthic triangle. So, we are done. $\square$


Solution 2

By Carnot's theorem, $AD,CF$ and the perpendicular from $E$ onto $AC$ are concurrent. Thus $AD,BE,CF$ are altitudes. It is well-known the altitudes of the triangle are the internal bisectors of its orthic triangle. So, we are done. $\square$

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