+965 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.
+1633 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.
+410 Thanks, I can confirm this is the correct answer, admist this confusion. 
We can verify proyaop's answer through similar triangles △BEF∼△ECD, so to check we make sure BFEF=EDCD, by proyaop's answer, BF=3−34=94, EF=34 , ED=34, CD=1−34=14, so 9434=3414=3.
Asinus did nothing majorly wrong, i think they misread the side lengths as 1 and 2 
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