Let $C$ be a point on line segment $AB$, and construct the circles with diameters $AB$ and $AC$. Let $M$ be the midpoint of $BC,$ and let $MD$ be a tangent to the smaller circle (with $D$ on the smaller circle). Point $E$ is on the larger circle such that $ME$ is perpendicular to $AB$, with $D$ and $E$ on the same side of $AB$. Prove that $A$, $D$, and $E$ are collinear.
We basically want to prove, since the center of homothety is at $A$, $C$ maps to $B$ and $D$ maps to $E$. We know $C$,$B$ and $A$ all lie on the same line, so $C$ does indeed maps to $B$.
Let $O_{1}$ and $O_{2}$ be the centers of the smaller circle and larger circle respectively. Since $O_{1}$ is mapped to $O_{2}$, it suffices to prove $\overline{DO_{1}}||\overline{EO_{2}}$.
We know $\overline{DO_{1}}\perp\overline{MD}$. We would thus like to show $\triangle{O_{1}DM}\cong\triangle{O_{2}ME}$ because, if we do so, then $\angle{MO_{1}D}=\angle{MO_{2}E}$ by Congruent Parts of Congruent Triangles are Congruent (CPCTC), which means $\overline{DO_{1}}||\overline{O_{2}E}$.
Since $D$ is a point of tangency, it must make right angles with the radius of the circle it's tangent to, so $\angle{O_{1}DM}=90^{\circ}$ and we're already given $\angle{EMO_{2}}=90^{\circ}$.
Now we only need another angle or a side length to finish the proof.
I need hints here, I know I'm yet to use the fact that $M$ is a midpoint of $CB$, but I don't know how.
Best Answer
Let $AE$ intersects the circle $O_1$ in the the points $A$ and $D'$.
Thus, $CD'||BE$ and $MCD'E$ is cyclic, which says $$\measuredangle CD'M=\measuredangle CEM=\measuredangle BEM=90^{\circ}-\measuredangle ABE=90^{\circ}-\measuredangle ACD'=\measuredangle CAD'.$$ Id est, $MD'$ is a tangent line to the circle $O_1$, which says $D\equiv D'$ and we are done!