+842 A square is inscribed in a right triangle, as shown below. The legs of the triangle are 1 and 3. Find the area of the square.
+1632 Through construction, AB = AF + FB = 3.
AF = EF = x. Since ADEF is a square, DE is parallel to AB, and thus angle CED is congruent to angle CBA by corresponding angles, and ADE = angle EFB because they are both 90 degrees. Thus, triangle CDE is similar to EFB is similar to CAB, and CA : AB = 1 : 3, likewise, FE : FB = 1 : 3. Substituting, we have x : FB = 1 : 3, and therefore FB = 3x. Substituting this into the original equation, AB = x + 3x = 4x = 3. And thus, x = 3/4. The area of the square would be x^2 = 9/16.
+399 Thanks, I can confirm this is the correct answer, admist this confusion. 
We can verify proyaop's answer through similar triangles \(\triangle BEF \sim\triangle ECD\), so to check we make sure \(\frac{BF}{EF}=\frac{ED}{CD}\), by proyaop's answer, \(BF=3-\frac{3}{4}=\frac{9}{4}\), \(EF=\frac{3}{4}\) , \(ED=\frac{3}{4}\), \(CD=1-\frac{3}{4}=\frac{1}{4}\), so \(\frac{\frac{9}{4}}{\frac{3}{4}}=\frac{\frac{3}{4}}{\frac{1}{4}}=3\).
Asinus did nothing majorly wrong, i think they misread the side lengths as 1 and 2 
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