Positioning Isosceles Triangle To Create Second Isosceles Triangle Two Triangles $ABC$ and $DEF$ are superimposed such that the points $D$ and $E$ fall on $AC$ and $BC$ respectively. $DEF$ is an isosceles triangle where $DF$ is equal to $EF$ !Problem Illustration Given the base and height of $DEF$ and the coordinates of points $(A_x,A_y)$, $(B_x,B_y)$, $(F_x,F_y)$: find $(C_x,C_y)$ such that $DEC$ is also an isosceles triangle. This seems like a relatively straight forward geometry problem, but I'm stumped. Any and all advice is welcome.

Let $M=(M_x,M_y)$ be the midpoint of $DE$: point $M$ belongs to the circle of center $F$ and given radius $FH=h$, so that $(M_x-F_x)^2+(M_y-F_y)^2=h^2$. Once $M$ is chosen on that circle, then the coordinates of points $D$ and $E$ can be found, because they are the intersections between the line through $M$ perpendicular to $FM$ and the circle of center $F$ and given radius $FD=FE=l$.

You can then write the equations of lines $AD$ and $BE$, as a function of $M_x$ and $M_y$: their common point $C$ must lie on line $FM$, and that condition allows you to find $M$ and thus $C$.