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I'm having trouble with this one:

Draw a parallelogram knowing the lengths of its sides and the angle between the diagonals.

Bonus points if the answer uses a translation, because that's where this exercise comes from.

I tried to find the length of the diagonals by using the cosine law, but I got a very complicated expression and I'm pretty sure there is a better way.

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    You might find the [Parallelogram Law](http://en.wikipedia.org/wiki/Parallelogram_law) useful, here.2012-12-06

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Fix one side, say segment $AB$, then draw (one arc of) the circle $c$ which consists of those points from where $AB$ is seen in the given angle. So that the centre of the parallelogramma has to be on $c$. Then enlarge $c$ to its double from center $A$, getting arc of circle $c'$ which must contain a $3$rd vertex of the parallelogramma, so finally just draw the circle centered at $B$ with the length of the other side.

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    $A$ and $B$ will be on $c$. Let $M$ be the midpoint of $AB$, and say, $\phi$ is the given angle. Then draw a right triangle over $AM$, having the right angle at $M$ and angle $90^\circ-\displaystyle\frac\phi2$ at $A$. Then we have $3$ points on $c$, draw the midperpendecular lines of these $3$ points and take their intersection.2012-12-08