How to inscribe an ellipse into an isosceles trapezoid? My main question is: how can I inscribe an ellipse into an isosceles trapezoid? I want to create an ellipse, whick is tanget to all four sides of the trapeziod (i.e. shares exactly one point with each).
Since I am not quite expert in mathematics, I would prefer going step by step.
First of all, is it even true, that a certain isosceles trapezoid determines only one ellipse that matches the critera above?
If yes, is it possible to construct it by euclidean construction (using straightedge and compass)?
In case it is not possible, is there any other way to create this ellipse? For example, with other construction tools?
Maybe if I put my trapezoid in a coordinate system, would it be possible to calculate the ellipse's focal points, or equation?
Searching for some solution on my own, I only could find this drawing:
https://www.geogebra.org/material/show/id/3075381
However, I am not skilled enough to decide it could be any help at all or not...
 A: 
Given 
$$ u,v, 2 b $$
Ellipse ( not sketched) touches slant line ; Calculate slope and distance to center $(m,h)$
$$ m= (v-u)/ (2 b) ,\quad h= b +u/m $$
Solve the two equns together
$$ ((x-h)/b)^2 + (y/A)^2 =1 , y = mx $$
$$  ( x^2 - 2 x h + h^2) +( m x b/A)^2 -b^2 = 0 $$
$$ x^2 ( 1+ (m b/A)^2) - 2 x h + ( h^2-b^2) = 0 $$
Point of tangency condition is for a double root or zero discriminant of quadratic equation. After simplification,
$$ A = m \sqrt{h^2-b^2} $$
Ellipse parametric equations are
$$ x = b\,\cos t +h, \quad y = A\, \sin t $$
The diagram below is constructed for numerical data
$$ ( u,v,b,m,h)=(3,5,2,0.5,8 ) $$

A: The isosceles trapezoid is a projection of a square onto the plane.
If a circle is inscribed in that square, the same projection
takes the circle to an ellipse inscribed in the isosceles trapezoid.
We can construct such a projection as follows:

On one base of the trapezoid, we construct a square.
The side that the square shares with the trapezoid is trivially projected
onto the same side of the trapezoid.
We extend lines $BG$ and $CF$ from the opposite base $BC$ of the trapezoid through the vertices of the opposite side $FG$ of the square, meeting at $H.$
Then $H$ is the center of the projection. 
A circle inscribed in the square touches the other two sides at $I$ and $J.$
The projection from $H$ projects $I$ to $E$ on one lateral side of the trapezoid and $J$ to $A$ on the other lateral side.
In fact, the projection from $H$ projects the entire circle inscribed in the
square onto an ellipse inscribed in the trapezoid
Since $A$ and $E$ are images of the points of tangency of the circle and square, they are themselves points of tangency of the ellipse and trapezoid.
The rest of the ellipse can be constructed using the two-circle construction as shown. One circle is the circle that touches the trapezoid at the points of tangency with the ellipse; this circle intersects $AE$ at $X,$ and then by extending $OX$ to intersect the line through $A$ perpendicular to $AE,$
and labeling the intersection point $Y,$ we find that $OY$ is the radius of the other circle in the construction of the ellipse.
We know the ellipse is not unique because we could have inscribed a non-circular ellipse in the square, and the image of that ellipse would again
be an ellipse inscribed in the trapezoid, but not the same one found in the construction above. However, the construction above gives the only ellipse that has mirror-image symmetry across the centerline of the trapezoid.
