Its a construction problem I am having trouble with. I realize I need to use rotations and/or other isometries but I am really stuck. Any help would be really appreciated!
Let the center of the circle be $O$. Rotate $B$ by $90^\circ$ about $O$ to get $B'$; that is, construct a point $B'$ such that $\lvert OB'\rvert = \lvert OB\rvert$ and $OB'\perp OB$. Then the problem is equivalent to constructing two chords of equal length through $A$ and $B'$ that are parallel to each other. In fact, these are the same chord: the one passing through both $A$ and $B'$. Rotate it by $-90^\circ$ about $O$ to get the chord through $B$ perpendicular to the chord through $A$.
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