$\triangle ABC$ is a triangle and A'B'C' are the midpoints of the sides $\overline{BC}, \overline{CA}$ and $\overline{AB}$ respectively. If $\overline{AD}$ is the altitude through $A$, prove that $\measuredangle B'DC'=\measuredangle B'A'C'$.
Hence. show that the circumcircle of $\Delta A'B'C'$ also passes through the feet $D, E, F$ of the altitudes of the $\Delta ABC$.
Given the midpoints A', B' and C', the triangles B'CA' and ACB are similar, which leads to B'A' = $\frac12$AB = AC'. Similarly, C'A' = AB'.
Given that AD $\perp$ BC and $C'$, $B'$ are the midpoints, $C'$ and $B'$ are the centers of the circumcircles for the right triangles ABD and ACD, respectively. Then, DC' = AC' = A'B' and DB' = AB' = A'C'. Along with the shared side C'B', the triangles A'B'C' and DB'C' are congruent, which yields $\angle$C'DB' = ∠C'A'B'.