If $OM=ON $ show that $EF=CD $.

Question

Let $\mathcal {C} $ a circle of center $O $.

Let $AB $ is diameter and and $EF ||CD $ with $ E,F,D,C \in \mathcal{C} $ s.t. $EF\cap AB=M $ and $CD\cap AB =N $.

If $OM=ON $ show that $EF=CD $.

I know that $EC=FD $.

Answer 1

Let $K\in EF$, $OK\perp EF$, $L\in CD$ and $OL\perp CD$.

Thus, since $EF||CD$, we obtain that $O\in KL$, $\Delta MKO\cong\Delta NLO,$ which gives $OK=OL$ and $EF=CD$.