Miguel Ochoa's van Schooten Like Theorem III
What Might This Be About?
In $\Delta ABC,$ circle $(O)$ through $A$ and tangent to $BC$ in $D$ cuts $AB$ and $AC$ in $E$ and $F,$ respectively.
Prove that $DE + DF = AD.$
In the notations I follow the original formulation:
Let $D'$ be the foot of the altitude from $A.$ Find $E'$ on $AC$ and $F'$ on $AB$ such that $DE'\perp AC$ and $DF'\perp AB.$
By Viviani's theorem, $AD'=DE'+DF'.$
Next we note that, since $AFDE$ is a cyclic quadrilateral, $\angle DFB=\angle DEA.$ Further, since the circle is tangent to $BC,$ $\angle ADB=\angle DEA.$ Thus, the right triangles $ADD',$ $DEE'$ and $DFF'$ are similar such that $AD'=DE'+DF'$ implies $AD=DE+DF.$