Show that a quadrilateral with perpendicular diagonals is circumscribing a circle

Question

A quadrilateral $ABCD$ is given with $AC \perp BD$ and $AB\cdot CD=AD\cdot BC$. Show that $ABCD$ is circumscribing a circle.

$AC\perp BD \Rightarrow AB^2+CD^2=BC^2+AD^2$ and we also have $AB\cdot CD=AD \cdot BC$. How to use the equalities to get that $AB+CD=AD+BC$?

Answer 1

Adding the first and twice the second equalities one obtains: $$ AB^2+2AB\cdot CD+CD^2=BC^2+2BC\cdot DA+DA^2\\ \implies (AB+CD)^2=(BC+DA)^2\\ \implies AB+CD=BC+DA. $$