a solid geometry problem

Question

In the following 3D figure, we know that $AE \bot EC, AD \bot BD$,

how to prove that $|ED| < |BC|$ ?

Answer 1

Choose $F$ such that $BDEF$ is a parallelogram. Now $DE$ is perpendicular to the plane (or degenerately line) through $C, E, F$, and $BF$ is parallel to $DE$, so $CFB$ is a right angle, so $|BC|^2 = |BF|^2 + |FC|^2 = |DE|^2 + |FC|^2 \geq |DE|^2$.