Let be the intersection of and , be the midpiont of , be the midpoint of , and be the intersection of and . We now claim that . First note that, since and [this is because ], we have that . Then , and , so . We earlier stated that , so we have that from SAS similarity. We have that , , and are collinear, and since and are on the same side of line , we can see that from SAS. Therefore , so , , and are collinear.
Now consider triangles and . Segments and are transversal lines, so it's not hard to see that . It's also not hard to show that by SAS similarity. Therefore , which implies that , , and are collinear. This completes the proof.
Define the same as above. Then there exists a positive homothety about mapping the homothety takes are collinear. Similarly, there is a negative homothety taking are collinear. So, are collinear, as desired .
Proof by IDMasterz