#### Answer

$A = \frac{ab+cd}{2}$

#### Work Step by Step

$s = \frac{a+b+c+d}{2}$
It is given in the question that $a^2+b^2 = c^2+d^2$
We can find an expression for the area of the cyclic quadrilateral:
$A = \sqrt{(s-a)(s-b)(s-c)(s-d)}$
$A = \sqrt{(\frac{a+b+c+d}{2}-a)(\frac{a+b+c+d}{2}-b)(\frac{a+b+c+d}{2}-c)(\frac{a+b+c+d}{2}-d)}$
$A = \sqrt{(\frac{b+c+d-a}{2})(\frac{a+c+d-b}{2})(\frac{a+b+d-c}{2})(\frac{a+b+c-d}{2})}$
$A = \sqrt{(\frac{-a^2-b^2+c^2+d^2+2ab+2cd}{4})(\frac{a^2+b^2-c^2-d^2+2ab+2cd}{4})}$
$A = \sqrt{(\frac{0+2ab+2cd}{4})(\frac{0+2ab+2cd}{4})}$
$A = \sqrt{(\frac{2ab+2cd}{4})(\frac{2ab+2cd}{4})}$
$A = \sqrt{(\frac{2ab+2cd}{4})^2}$
$A = \frac{2ab+2cd}{4}$
$A = \frac{ab+cd}{2}$