This is a problem from the Regional Mathematics Olympiad, RMO 2015 Mumbai Region based on Diagonal of a Quadrilateral. Try to solve it.
Problem: Diagonal of a Quadrilateral
Let ABCD be a convex quadrilateral with AB = a, BC = b, CD = c and DA = d. Suppose $ a^2 + b^2 + c^2 + d^2 = ab + bc + cd + da $ and the area of ABCD is 60 square units. If the length of one of the diagonals is 30 unit, determine the length of the other diagonal.
Discussion:
(Solution suggested in class by Megha Chakraborty)
It is given that $ a^2 + b^2 + c^2 + d^2 = ab + bc + cd + da $ and $ s=2$
Multiplying 2 to both sides we have
$ 2a^2 + 2b^2 + 2c^2 + 2d^2 = 2ab + 2bc + 2cd + 2da $ and $ s=2$
$ \Rightarrow a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2cd + d^2 + d^2 - 2ad + a^2 = 0 $ and $ s=2$
$ \Rightarrow (a-b)^2 + (b-c)^2 + (c-d)^2 + (d-a)^2 = 0 $ and $ s=2$
But sum of squares can be 0 if and only if each square is individually 0. This implies a = b = c = d. Hence the quadrilateral is a rhombus (in a special case, a square).
The area of a rhombus is $ \frac{1}{2}\times{d_1}\times {d_2} $ and $ s=2$ where $ d_1 $ and $ s=2 $ and $ d_2 $ and $ s=2 $ are the diagonals. It is given that one of the diagonals is 30 and area is 60. Hence we have
$ \frac{1}{2}\times 30\times {d_2} =60 $ and $ s=2$
$ \Rightarrow {d_2} = 4 $ and $ s=2$
Hence the length of the other diagonal is 4.
