euclidean-geometrygeometry
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.
This is true by symmetry. The figure is symmetric with respect to reflection in the line joing the centres; if the line joining the two intersection points weren't perpendicular to that line, it would break the symmetry.
Hint:
$ACSR$ is a rectangle. Do you see why $AC=RS=\frac12 LM$?
Solution: construct - as was suggested - a right triangle $ABC$ whose hypotenuse $AB$ is the segment between the centers of the given disks, and one of the legs ($AC$) is congruent to $\frac12EF$. Draw through the point $K$ a line parallel to $AC$. It will intersect the circles in the points $L$ and $M$. Then $LM=EF$. The construction is possible only if $\frac12EF<AB$, in which case there are two solutions according to the number of possibilities to construct the triangle $ABC$.
Best Answer
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 $AB = CD$.