Sixteenth STARS OF MATHEMATICS Competition

Notice that $\angle PDC + \angle PBC = \angle PDA + \angle PBA$ implies $\angle PDA + \angle PBA = (B + D)/2$. In the same manner, $\angle PAB + \angle PCB = (A + C)/2$. Add the last equalities to obtain $180^\circ = (A + B + C + D)/2 = (\angle PBA + \angle PAB) + \angle PDA + \angle PCB = (180^\circ - \angle APB) + \angle PDA + \angle PCB$, and notice that $\angle PDA + \angle PCB = \angle APB$ shows that circles $APD$ and $BPC$ are tangent at $P$.

Let $(O_1, R_1)$, $(O_2, R_2)$, $(O_3, R_3)$, $(O_4, R_4)$ be the circles $PAB$, $PBC$, $PCD$ and $PDA$ respectively. Recall that points $P$, $O_1$, $O_3$ are collinear, and similarly, points $P$, $O_2$, $O_4$ are collinear. Further, $\angle O_2O_1O_4 + \angle O_2O_3O_4 = 180^\circ - \angle APB + 180^\circ - \angle CPD = 180^\circ$, so $O_1O_2O_3O_4$ is a cyclic quadrilateral; let $R$ be its circumradius.

By Sine Law,
$$2R = \frac{O_1O_3}{\sin O_2} = \frac{O_2O_4}{\sin O_1} \implies \frac{R_1 + R_3}{\sin \angle BPC} = \frac{R_2 + R_4}{\sin \angle APB},$$
then $\frac{AB}{\sin \angle APB} + \frac{CD}{\sin \angle CPD} = 2(R_1 + R_3) \implies AB + CD = 2(R_1 + R_3) \sin \angle APB$. Similarly, $BC + DA = 2(R_2 + R_4) \sin \angle BPC$. All the above lead to $AB + CD = BC + DA$, hence the claim.