When an ellipse touches the sides of a triangle An ellipse touches the sides of a triangle $abc$ from inside in the points $a',b',c'$.
How can I prove, that the lines $ aa',bb',cc'$ meet in one point?
The ellipse equation is : $ \frac{x^2}{A^2} + \frac{x^2}{B^2}= 1 $
I've seen this kind of questions in old exams, therefore I would like to know how to handle such a proof right.
Do I have to build equations for $aa',bb',cc'$ ? If yes, how do do that?
 A: EDIT: the proof below works only for Steiner inscribed ellipse, not for any inscribed ellipse. Which makes this proof useless, sorry for that.
Depending the date of the "old exams" where you have seen this problem, it may have been solved by using affine transformations. Two people have already commented about that, so here are more details.
The construction is invariant by affine transformations, because it involves only operations that have this invariant: tangents, intersections. A triangle is transformed into a triangle, an ellipse is transformed into an ellipse, and as the ellipse inscribed into a triangle is unique, if a triangle T is transformed into a triangle U the ellipse inscribed in T gets transformed into the ellipse inscribed in U.
Then, there is an affine transformation that transforms the triangle $abc$ into a given equilateral triangle. So the intersection of $aa'$ with $bb'$ gets transformed into the intersection of $aa'$ with $bb'$ in the equilateral triangle, etc.
Proving the property for an equilateral triangle is easy, because the inscribed ellipse is a circle and the 3 lines intersect at the circle center. So the fact that the 3 lines intersect on only 1 point gets transformed back into the same property for the first triangle, because the number of intersections is an affine invariant. Proof is finished. And you don't have to explicit the equations.
It may seem handwavy, but given some geometry background that was current some 80 (?) years ago, it is perfectly valid. And the nice thing is, it applies to many constructions, especially with triangles and parallelograms.
A: A much simpler answer is obtained based on the fact that any triangle with any inscribed ellipse, can be transformed to a triangle where the inscribed ellipse is transformed into a circle.  This circle, of course, is the incircle.  It can be shown easily that the cevians (the line segments) joining the vertices to the opposites sides touching points of the incircle are concurrent, based on the properties of the incircle, and Ceva's theorem.  A very to-the-point proof is given here.
A: Lateral comment only, not an answer.
$$(x_c,y_c)=\left(\frac{t_1- t_1t_2}{1-t_1t_2},\frac{t_2-t_1 t_2}{1-t_1t_2}\right)$$

