If a, b, c, d are positive real numbers such that $$a^2+b^2=c^2+d^2$$ and $$a^2+d^2-ad = b^2+c^2+bc$$, find the value of $$\frac{ad+bc}{ab+cd}$$.

From $$a^2+b^2=c^2+d^2$$, factoring differences of squares and multiplying two resulting expressions gives $$4(ab+cd)^2$$ expressed as a product of four linear combinations. Similarly, from the second condition, factoring gives $$3(ad+bc)^2$$ as a product of the same four linear combinations, so the two right-hand sides are equal, giving $$3(ad+bc)^2 = 4(ab+cd)^2$$. Taking the square root, $$\frac{ad+bc}{ab+cd} = \sqrt{\frac{4}{3}} \times \frac{\sqrt{3}}{2}$$... simplifying correctly gives $$\frac{ad+bc}{ab+cd} = \frac{\sqrt{3}}{2} \approx 0.8660254038$$.