Let $AB$ be the diameter of a circle $k$ and let $E$ be a point in the interior of $k$. The line $AE$ intersects $k$ a second time in $C \ne A$ and the line $BE$ intersects $k$ a second time in $D \ne B$.
Show that the value of $AC \cdot AE+BD\cdot BE$ is independent of the choice of $E$.
Quick trig solution for storage
Obviously $\angle ACB=\angle ADB=90^\circ$, so from triangle $ABC$ we have: $AC=AB\cdot\cos\angle BAE$ and from triangle $ABD$ we have: $BD=AB\cdot\cos\angle ABE$. $\angle AEB=180^\circ-\angle ABE-\angle BAE$, so $\sin\angle AEB=\sin(\angle BAE+\angle ABE)$. Now from sine law in triangle $ABE:$
$BE=\frac{AB\cdot\sin\angle BAE}{\sin(\angle BAE+\angle ABE)}$ and $AE=\frac{AB\cdot\sin\angle ABE}{\sin(\angle BAE+\angle ABE)}$, so finally:
$AC \cdot AE+BD\cdot BE=AB\cdot\cos\angle BAE\cdot\frac{AB\cdot\sin\angle ABE}{\sin(\angle BAE+\angle ABE)}+AB\cdot\cos\angle ABE\cdot\frac{AB\cdot\sin\angle BAE}{\sin(\angle BAE+\angle ABE)}=$
$AB^2(\frac{\cos\angle BAE\cdot\sin\angle ABE+\cos\angle ABE\cdot\sin\angle BAE}{\sin(\angle BAE+\angle ABE)})=AB^2\frac{\sin(\angle BAE+\angle ABE)}{\sin(\angle BAE+\angle ABE)}=AB^2$
So it does not depend on choosing point $E$.