Tangent circumferences to the sides of a parallelogram with two fixed points in its diagonal I'm asking about this problem:
Let a parallelogram ABCD, and two points E and F in the diagonal AC, if it exists a circumference by E and F tangent to the sides AB and BC then it exists another circumference in E and F tangent to the sides (or in its prolongation) DA and DC.
I conjecture that the tangents points of this circumference are the intersection of the prolongation of the sides DA, DC and the straight line throw U and V who are the points of tangence of the former circumference in the sides AB and BC. But I'm unable to prove that.
I draw this picture
 A: Draw line perpendicular to $DT$ at point $T$ and line perpendicular to $DS$ at point $S$. These two lines intersect at point $O$ and we should prove that it is the center of the big circle passing through points $E$ and $F$.
First, we have to prove that $OT=OS$. Notice that $BU=BV$ so triangle $BUV$ is isosceles. It's easy to see that triangles $CVT$ and $ASU$ are isosceles too. It means that:
$$CV=CT,\quad AU=AS\tag{1}$$
It also means that all acute angles of triangles $ASU$, $BUV$, $CVT$ are equal. In particular:
$$\angle ASU=\angle CTV$$
...and because of that:
$$\angle TSO=\angle STO$$
In other words, triangle $OST$ is isosceles and $OT$=$OS$. So there is a circle with center $O$ and radius $OS=OT$.
Now take a look at (1). Potential of points $A,C$ with respect to small and big circles are equal. It means that points $A$ and $C$ must be on the radical axis $AC$ of two circles. Radical axis $AC$ passes through intersection points of two circles. In other words, the big circle must pass through points of intersection of radical axis $AC$ and small circle. And these two points are points $E,F$. So the big circle has to pass through points $E,F$ as well. This completes the proof.
