I need help with the following triangle similarity problem:
A rectangle's diagonal is divided into thirds by heights from two opposite points of the rectangle. If the length of one of the sides of the rectangle is $\sqrt{2}$, find the other side.
If you look at the photo, the heights divide the rectangle into 4 right triangles. I have proved triangles DEC and AFB, and the smaller two, DEA and BFC to be similar, but cannot prove one of for example triangle DEC and DEA to be similar. But even if I could, I am not sure what to do next to find the other side of the rectangle, since setting up any kind of proportion did not seem to lead to the solution.
Any help would be appreciated!
To prove similarity you need to show that corresponding angles are equal. You have two tools at your disposal:
Label one of the non-right angles as $\alpha$, and you should be able to use the above tools to label others as either $(90°-\alpha)$ or as $\alpha$ again. That should give you the similarity you were searching for.
If all these triangles are similar, then there is a constant ratio between the long and the short leg of each. Let's call that factor $f$. If you multiply the short leg of a small triangle (like $\triangle DEA$) by $f$ your get the long leg. For example $\overline{EA}\cdot f=\overline{ED}$. Likewise if you multiply the short leg of a large triangle (like $\triangle DEC$) by $f$ you get the long leg. $\overline{ED}\cdot f=\overline{EC}$.
Now note that the long lef of a small triangle is the short leg of a large triangle. $\overline{ED}$ occurs in both the formulas above. Do you can combine them to $\overline{EA}\cdot f^2=\overline{EC}$. But that ratio is along the diagonal, so you can conclude $f^2=2$ and from that drive $f$. Since that is not only the ratio between the legs of a single triangle, but also the ratio between all the edges of a large triangle and the corresponding edges of a small triangle, it is the ratio you need to answer the question.