It's easy to show that the horizontal component of $M$ is $6$, and the vertical component is $h/2$. 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 $$h^2=18$$

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 $$h^2=18$$