Chords and the Center

Let's say I have a circle $\omega$ with center $O$.

And a chord $AB$.

Now make a perpendicular line from $O$ to $AB$

You've guessed it! If we have a chord $AB$ in a circle, and we draw a perpendicular line from $O$ to the chord, then this perpendicular line bisects $AB$

But is the converse true?

So how do we prove the first statement? Think about it; it's actually quite simple :)

Because $OA = OB$ and $OC = OC$, we have $\triangle AOC \cong \triangle BOC$ by HL congruence. Now, try to prove the converse on your own!