Given a circle (O, R) with diameter AB. Point M on (O), A, B are not coincident. Two lines through O and perpendicular to AM, BM intersects the tangent of (O) through M at C, D, respectively. OC intersects AM at I, OD intersects BM at K. Prove that IK, AD, BC are concurrent.
Attempts: I tried drawing an altitude through M of triangle ABC, intersecting IK at some point but still stuck on proving that it is the midpoint of that altitude. AC, BD are tangents of (O) and I, K are midpoints of AM, BM respectively has been proved.
Best Answer
Let $IK$ intersects $AD$ at $S$. Then homothety at $S$ which takes $A$ to $D$ takes $I$ to $K$. If we prove that it takes also $C$ to $B$ then we are done.
Now $CA =CM$ so the line $CA$ is also tangent to (semi) cricle at $A$ and thus $CA\bot AB$. The same way we have $BD\bot AB$, so $CA||BD$. But this now means that $C$ goes to $B$ and we are done.