socrates wrote: Given positive real numbers $a, b, c, d$ that satisfy equalities \[a^2 + d^2 - ad = b^2 + c^2 + bc \ \ \text{and} \ \ a^2 + b^2 = c^2 + d^2\] find all possible values of the expression $\frac{ab+cd}{ad+bc}.$ We get ($S=a+b+c+d$): $a^2+d^2-ad+(bc-ad)=b^2+c^2+bc+(bc-ad)\Rightarrow bc+ad=(S-2a)(S-2d)$ $a^2+d^2-ad+(3ad-3bc)=b^2+c^2+bc+(3ad-3bc)$ $\Rightarrow 3(bc+ad)=(S-2b)(S-2c)$ So $3(bc+ad)^2=(S-2a)(S-2b)(S-2c)(S-2d)$. And we get: $a^2+b^2+(2ab-2cd)=c^2+d^2+(2ab-2cd)\Rightarrow 2(ab+cd)=(S-2c)(S-2d)$ $a^2+b^2+(2cd-2ab)=c^2+d^2+(2cd-2ab)\Rightarrow 2(ab+cd)=(S-2a)(S-2b)$ So $4(ab+cd)^2=(S-2a)(S-2b)(S-2c)(S-2d)$. Hence $\frac{ab+cd}{ad+bc}=\frac{\sqrt{3}}{2}$.

$ad+bc=a^2-c^2+d^2-b^2=2(a^2-c^2)=2(d^2-b^2)$, $\frac{ab+cd}{ad+bc}=\sqrt{1-\frac{(a^2-c^2)(d^2-b^2)}{(ad+bc)^2}}=\frac{\sqrt{3}}{2}.$

We also get as a nice geometric solution that there exists a cyclic quadrilateral $ABCD$ with sidelengths $AB=a, BC=b, CD=c$ and $DA=d$ which has the angles $CBA=ADC=90^\circ$ and $BAD=60^\circ$. Now, $ab+cd$ is just half of the area of this quadrilateral and $ad+bc$ is $\frac{1}{2 \sin 60^\circ}$ times the area, therefore their quotient establishes to be $\sin 60^\circ=\frac{\sqrt{3}}{2}$.

socrates wrote: Given positive real numbers $a, b, c, d$ that satisfy equalities \[a^2 + d^2 - ad = b^2 + c^2 + bc \ \ \text{and} \ \ a^2 + b^2 = c^2 + d^2\]find all possible values of the expression $\frac{ab+cd}{ad+bc}.$ NMTC 2017 Juniors P6 http://mathometer.weebly.com/uploads/7/1/3/5/71354423/rmo_mock_1.pdf