A chord $AB$ of one of two concentric circles at intersect each other at $C$ and $D$. We have to prove, $AC=BD$.
I am not sure what this question means by 'intersect each other', but if I am correct, we can assume that $AB$ is the chord of the outer circle that intersects the inner one at $C$ and $D$. I proved in one of my exams that if their centre is at $O$, triangles $\triangle ACO$ and $\triangle BDO$ are congruent by SAS. Thus $AC=BD$, proving the theorem. The teacher, however, gave me zero marks and left only one comment, 'Wrong derivation'. I do not even know what he means.So what is wrong with my proof?
ADDED: For a more detailed explanation of what I did, I joined $O$ with $A$, $B$, $C$ and $D$. I argued that since $OC$ and $OD$ are equal, angles $\angle OCD$ and $\angle ODA$ are equal, and thus angles $\angle OCA$ and $\angle ODB$ are equal. By similar reasoning, $\angle OAC$ and $\angle OBD$ must be equal. Therefore, the remaining angles must be equal. Therefore, by SAS, triangles $\triangle ACO$ and $\triangle DBO$ are congruent. Thus $AC$ and $BD$ are equal.