A Property of Nagel Cevians


A Property of Nagel Cevians, source


A Property of Nagel Cevians, problem


Let $E$ be the point of tangecy of the incircle with $BC.$ Then $FE$ is a diameter of the incircle.

Indeed, draw the excircle $(G)$ opposite $A.$ By definition, $D$ is the point of tangency of $BC$ and $(G).$ If $I$ is the incenter and $(I)$ the incircle, $(G)$ is homothetic to $(I)$ from $A$ with the coefficient $AD/AF.$ Note that $GD\perp BC.$

A Property of Nagel Cevians, lemma

If $DJ$ is a diameter of $(G)$ then $J$ is the image of $E$ under the homothety. So, it follows that $FE\perp BC,$ making $FE$ a diameter.


Let $s$ be the semiperimeter of $\Delta ABC$ with the side lengths $a,b,c.$ Introduce $x=s-a,$ $y=s-b,$ $z=s-c.$ Then $a=y+z,$ $b=z+x,$ $c=x+y.$ Since the barycentric coordinates of $I$ are $(a:b:c)=(y+z:z+x:x+y)$ and those of the Nagel point $(s-a:s-b:s-c)=(x:y:z),$ we have (using the above lemma, too):

$\displaystyle\begin{align} \overset{\rightarrow}{AE}&=\frac{z\overset{\rightarrow}{AB}+y\overset{\rightarrow}{AC}}{y+z}\\ \overset{\rightarrow}{AN_a}&=\frac{y\overset{\rightarrow}{AB}+z\overset{\rightarrow}{AC}}{x+y+z}\\ 2\overset{\rightarrow}{AI}&=\frac{(x+z)\overset{\rightarrow}{AB}+(x+y)\overset{\rightarrow}{AC}}{x+y+z}\\ \overset{\rightarrow}{AD}&=\frac{y\overset{\rightarrow}{AB}+z\overset{\rightarrow}{AC}}{y+z}\\ \end{align}$

Since $I$ is the midpoint of $FE$,


It follows that $\displaystyle\frac{AN_a}{AF}=\frac{y+z}{x}.$

But, from the van Aubel theorem, $\displaystyle\frac{AN_a}{N_aD}=\frac{y+z}{x}.$

This completes the proof.


The problem (by Kadir Altintas) has been posted at the Peru Geometrico facebook group and kindly communicated to me by Leo Giugiuc, along with a solution of his. Additional solutions can be found at the above link.


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