Consider $\Delta ABC$ has $AB=AC\text{ and } \angle BAC=90^o$. Let $O$ with radius $\dfrac{AB}2$. The secant $DE$ such that $CE$ always goes through $OB$. $CO \cap BD=M;CO \cap BE= N$. Prove that $OM=ON$
To prove $OM=ON$ we need prove $MANB$ is parallelogram $(OA=OB)$
We need to prove $\angle MAB=\angle ABN\rightarrow \angle MAC=\angle BAE=\angle BDE=\angle CDM$
I dont know how to prove this. I tried to prove some similar triangles but failed.
hint: if you can approve TE pass O , then the problem is solved.