Prove that IK, AC, BD are concurrent.

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.

• Where are you stuck? Show your attempts. – cosmo5 Oct 31 '20 at 4:18
• This is a direct consequence of Pappus' theorem, applied for lines $AOB$ and $CDM$. – lminsl Oct 31 '20 at 9:48
• Is analytical geometry allowed? – Oldboy Oct 31 '20 at 12:56
• Analytical geometry is not allowed. Also, Pappus' theorem is way too advanced for this problem, and I'm not allowed to use it. – Triet Vo Nguyen Minh Oct 31 '20 at 14:00
• Take this with a large grain of salt, given that I am generally ignorant of Geometry. "Also, Pappus' theorem is way too advanced for this problem" Sometimes, in a case like this, if all else fails, you can (for example) examine the proof to Pappus' theorem and see if you can "lift out some of the analysis" and use it to reverse-engineer a simple proof to your query. – user2661923 Oct 31 '20 at 19:40

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.