Problem 1903

Solution:

Let $O$ be center of $k$. We deduce that $B P = B Q$, $C Q = C R$, $D R = D S$, since those are tangents to the circle $k$. Using the condition $A B + C D = B C + D E$, we derive:
$$A P + B P + C R + D R = B Q + C Q + D S + E S$$
From here we have $A P = E S$.
Thus,
$$\triangle A P O \cong \triangle E S O \left(A P = E S, \angle A P O = \angle E S O = 90^{\circ}, P O = S O\right)$$
This implies
$$\angle O P S = \angle O S P$$
Therefore,
$$\angle A P S = \angle A P O + \angle O P S = 90^{\circ} + \angle O P S = 90^{\circ} + \angle O S P = \angle P S E$$
Now, from quadrilateral $A P S E$ we deduce:
$$2 \angle E A P + 2 \angle A P S = \angle E A P + \angle A P S + \angle P S E + \angle S E A = 360^{\circ}$$
So,
$$\angle E A P + \angle A P S = 180^{\circ}$$
and $A P S E$ is isosceles trapezoid. Therefore, $A E \| P S$.