# Cevians and Conic through Six Points on Sidelines of Triangle

### What is this about?

7 July 2013, Created with GeoGebra

### Problem

In $\Delta ABC$ cevians $AD$, $BE$, and $CF$ meet at point $W$.

Prove that there is a conic tangent to $BC$ at $D,$ $AC$ at $E,$ and $AB$ at $F.$

### Hint

If the statement is correct, and there is indeed a conic through the six points, you can change the configuration to a more convenient one by an affine transformation without disturbing its essential properties: the cevians remain concurrent, the points on each side remain equidistant from the feet of the cevians, the ellipse passes through six points.

### Solution

To start off, I'll use the same approach that worked so well in case where points $D,E,F$ were the midpoints of the sides of $\Delta ABC.$ Make an affine transformation that maps $\Delta ABC$ on to the right isosceles triangles with vertices $(0,0),$ $(1,0),$ $(0,1),$ with points $D,F,E$ at $(m,0),$ $(0,n),$ and $(u,v),$ respecively; $u+v=1.$

By Ceva's theorem, $\frac{1-n}{n}\cdot\frac{m}{1-m}\cdot\frac{v}{1-v}=1,$ or, which is the same, $\frac{1-n}{n}\cdot\frac{m}{1-m}\cdot\frac{1-u}{u}=1.$ Thus, we can express $u$ and $v$ in terms of $m$ and $n:$

$u=\frac{mn}{(1-m)(1-n)+mn}$, $v=\frac{(1-m)(1-n)}{(1-m)(1-n)+mn}.$

As before, we are going to look for a conic (if such exists) through the six points. A conic is given by a second degree equation:

$ax^{2}+2bxy+cy^{2}+2dx+2ey+f=0,$

with only $5$ coefficients independent (because the conic would not change after the equation was divided by one of non-zero coefficients.)

Let's substitute the coordinates of points $G,H,I,J,K,L$ into the equation:

$a(m\pm h)^{2}+2d(m\pm h)+f=0,$

$c(n\pm i)^{2}+2e(n\pm i)+f=0,$

$a(u\pm j)^{2}+2b(u\pm j)(v\mp j)+c(v\mp j)^{2}+2d(u\pm j)+2e(v\mp j)+f=0.$

Unfold the first equation into two:

$a(m^{2}+2mh+h^{2})+2d(m+h)+f=0,$ and

$a(m^{2}-2mh+h^{2})+2d(m-h)+f=0.$

Solving these for $a$ and $d$ gives

$a = f/(h^{2}-m^{2}), $

$d = -ma.$

Similarly, unfolding the second pair of equations, we obtain

$c = f/(i^{2}-n^{2}), $

$e = -nc.$

Adding and subtracting the two unfolded equations for $I$ and $J$ gives

$a(u^{2}+j^{2})+2b(uv-j^{2})+c(v^{2}+j^{2})+2ud+2ve+f=0,$ and

$aj-cj+dj-ej=0,$

the latter of which holds automatically because, as we found, $a+d=c+e=0.$ From the former we can find the remaining coefficient $b$ in terms of $f.$ The six points $G,H,I,J,K,L$ determine a unique ellipse.

### Acknowledgment

This problem generalizes the previously considered case where points $D,E,F$ were the midpoints of the sides of $\Delta ABC.$ The current problem, as the previous one, has been suggested by Dao Thanh Oai at the CutTheKnotMath facebook page.

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