$d_1^2\cdot d_2^2=a^2\cdot c^2+b^2\cdot d^2$ iff $A+C=\frac{\pi}{2}$ or $B+D=\frac{\pi}{2}$.

Question

Let $ABCD$ a quadrilateral, $a,b,c,d$ the lengths of his edges and $d_1,d_2$ the lengths of diagonals. Show that $d_1^2\cdot d_2^2=a^2\cdot c^2+b^2\cdot d^2$ iff $A+C=\frac{\pi}{2}$ or $B+D=\frac{\pi}{2}$.

I tried to prove it with scalar product of vectors but I am stuck.

My idea: $d_1^2=\vec{AC}^2=(\vec{AB}+\vec{BC})^2=AB^2+2 \vec{AB}\cdot \vec{BC} +{BC}^2$.

$d_2^2=\vec{BD}^2=(\vec{BC}+\vec{CD})^2=BC^2+2 \vec{BC}\cdot \vec{CD} +{CD}^2$.

Answer 1

I have checked the sufficiency by algebra, but there is more synthetic treatment, e.g. see this blog.