Prove that if through an intersection point of two circles, we draw all secant segments without extending them to the exterior of the disks, then the greatest of these secants will be the one parallel to the line of centers.
In the diagram below, $I$ is the intersection of the two circles and $EF$ is the line of centers. We are asked to prove that $DC$, parallel to $EF$, is longer than any other line $AB$ through $I$.
EDIT: Added point $J$ to the diagram.
All the triangles $ABJ$ where $J$ is second intersection, are similary, so $AB$ will be the longest iff $AJ$ will be the longest, that is iff $AJ$ is diameter of left (and $BJ$ diameter of right) circle.
So you are correct, in that case $\angle AIJ = 90^{\circ}$ so $AB = CD$.