Do the tangent points of the circles that are tangent to the sides of a triangle lie on a conic section?

Question

I came up with this feature about 3 months ago while using GeoGebra

If we have a triangle, and we draw the three circles that touch its sides from the outside, then the six points of contact lie on one conic section. Is this property already known? How do we prove it anyway?

Answer 1

Let us consider the following diagram: We know that the tangents to a circle drawn from a point outside it are of equal length, therefore: $AQ=AT$

$⇒AB+BQ=AC+CT\cdots [1]$

$BQ=BD\cdots [2]$

$CT=CD\cdots [3]$

$[1],[2],[3]⇒AB+BD=AC+CD=\frac{AB+BC+CA}{2}=S$

$⇒AQ=AT=S$

In the same way we can find:

$AQ=AT=BN=BV=CM=CP=S$

$AN=BQ,CV=BP,AM=CT$

Now we can apply Carnot's theorem:

$\frac{AQ}{BQ}\cdot\frac{AN}{BN}\cdot\frac{BV}{CV}\cdot\frac{BP}{CP}\cdot\frac{CM}{AM}\cdot\frac{CT}{AT}=1$

Hence the points $M,N,P,Q,T,V$ lie on a conic section

Answer 2

Now when I looked at the Carnot's Theorem page on the Cutting the Knot website and looked at the related topics, I found that my theorem had already been discovered.

https://www.cut-the-knot.org/m/Geometry/InExConics.shtml