Trapezoid height from one base to other base given tangent circle

Question

In trapezoid $ABCD$ with $AB \parallel CD$ and $AB=3, CD=7,$ and $AD=BC,$ define $M$ to be the midpoint of side $BC.$ The circle with diameter $DM$ is tangent to line $AB.$ What is the length of the altitude from $AB$ to $CD$?

I started by drawing a roughly accurate diagram on geogebra.

It seems that the point of tangency of the circle with diameter $DM$ to $AB$ is a trisection point, but I'm not sure. I also drew $MG$ the midline of $ABCD.$ Then if $h$ is the height from $A$ to $MG$ the area of $ABCD$ is $4h+6h=10h.$ Then it suffices to find the area of $ABCD.$ But I'm not sure how to do this. Furthermore I'm not sure how to relate the circle with diameter $DM$ to the problem, all I know is that if $F$ is the center then $\angle FEA=90$ and $\angle DEM = 90.$ May I have some help? Thanks in advance.

Answer 1

It's easy to show that the horizontal component of $M$ is $6$, and the vertical component is $h/2$ ($h$ is the height $AH$). Then $$DM=\sqrt{6^2+\frac{h^2}4}$$ Also easy to see that the vertical position of $F$ is $h/4$, so the radius of the circle is $3h/4$. Then $DM $ is twice the radius. $$\sqrt{6^2+\frac{h^2}4}=\frac32h$$ Squaring it and separating $h^2$, you get $$h^2=18$$

Answer 2

I am using $h$ as defined by the OP.

The altitude of $M$ to $DC$ is also $h$ and has base point $I$ on $DC$. As $F$ is the midpoint to $DM$ using scaling the altitude of $F$ to $DC$ is $h/2$. Thus $EF=3/2\cdot h$. $EF$ is a radius and $DM$ a diameter, thus $DM=2EF=3h$.

Because $AD=BC$ your trapezoid is symmetric. Thus $CI=((DC-AB)/2)/2=1$ and $DI=DC-CI=6$.

Apply pythagoras to $DIM$: $6^2+h^2=(3h)^2 \Rightarrow h=\frac{3}{\sqrt{2}}$.

The altitude is $2h=3\sqrt{2}$.