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Given $ABCD$ a rectangular trapezoid, $\angle A=90^\circ$, $AB\parallel DC$, $2AB = CD$ and $AC \perp BD$. What is the value of $AC/BD$ ?

Attempts so far: I have tried using the ratio of the areas of triangles $AOB$ and $DOC$, which is $\frac14$ (where $O$ is the intersection of the diagonals), but I couldn't get anything useful. I don't know how to use the fact that the diagonals are perpendicular.

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  • $\begingroup$ Drop and altitude and turn the trapezoid into a rectangle and a triangle. $\endgroup$
    – Doug M
    Jul 13 '16 at 21:13
  • $\begingroup$ I have tried to use the ratio of the areas of triangles AOB and DOC, which is 1/4 (where O is the intersection of the diagonals), but I couldn't get anything useful. I don't know how to use the fact that the diagonals are perpendicular. $\endgroup$
    – lucianzr1
    Jul 13 '16 at 21:15
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Put $AB=1$, $DC=2$, $AD=x$, and set the trapezoid with $A$ in the origin of a cartesian plane. Take the vector $AC=(x,2)$ and the vector $DB=(-x,1)$ and impose that they be orthogonal, i.e. their dot product be null, i.e. $x^2=2$. Now you have all the data to continue.

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