Question 169
The diagonals of convex quadrilateral $ABCD$ meet at point $E$. Given
$$
\frac{|AB|}{|CD|} = \frac{|BC|}{|AD|} = \sqrt{\frac{|BE|}{|DE|}}
$$
show that $ABCD$ is either a parallelogram or a cyclic quadrilateral.
$$
\frac{|AB|}{|CD|} = \frac{|BC|}{|AD|} = \sqrt{\frac{|BE|}{|DE|}}
$$
show that $ABCD$ is either a parallelogram or a cyclic quadrilateral.