square inscribed in a right triangle     A square of maximum possible area is circumscribed by a right angle triangle ABC
    in such a way that one of its side just lies on the hypotenuse of the triangle.
    What is the area of the square?
    actually the answer is given as
 $(abc/(a^2+b^2+ab))^2$
    Please provide the approach to solve the problem.
 A: Hint:
$\hspace{60pt}$
The red solid line is the height dropped onto the hypotenuse, i.e. $h = \frac{ab}{c}$ and the red dotted lines are of the same length. The green parallel lines are unnecessary, but might get you some intuitions.
Good luck! ;-)
A: Consider - a,b as right legs and c as the hypotenuse.
Let side of square = s
AC = b, BC = a, AB = c.

FB = as/b and AE = bs/a as the colored triangles are similar to the bigger triangle.
Steps to calculate area (S^2) :
1)Calculate GB and AD using right angle triangle rule for triangles GBF and ADE.
2)Calculate GD using right angle triangle rule for triangle GCD.
3)GD^2 = s^2. You get a quadratic equation in s which can be factorized. You get s = (abc)/(a^2 + b^2 +a.b)
If still not clarified will post the answer then.
A: Hint: 
The square formed some similar right-angled triangles, make use of the ratio of the sides.
A: The question can be easily solved by subtracting the two sides of a right triangle from the bigger triangle
i.e. subtract 'as/b' and 'bs/a' from the hypotenuse of larger triangle 'c'.and it will be equal to the side of a square.
S=c-(as/b+bs/a).
And thus s=abc/a^2+b^2+ab
