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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!

Thanks!

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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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Can you please make it clear by what you mean "Then it is equivalent..." what is equivalent to what? –  UH1 Dec 12 '12 at 4:44
    
I guess i understand that you are saying the problem is equivalent to constructing two chords of equal length through A and B' that are parallel to each other but I don't know how to solve that either (i think its equally hard) –  UH1 Dec 12 '12 at 5:05
    
@UH1: "In fact, these are the same chord: the one passing through both $A$ and $B'$." –  Rahul Dec 12 '12 at 5:34
    
are you suggesting that the chord passing through A and B' is the same length as the two chords of equal length that are parallel and pass through A and B?? –  UH1 Dec 12 '12 at 5:37
    
If you know how to construct two such chords (one passing through and the other through B') that they are equal in length and parallel to each other, please describe the solution. thanks! –  UH1 Dec 12 '12 at 5:46

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