Problem 2090

Solution:
We shall use the following fact: if $\alpha, \beta, \gamma$ and $\delta$ are angles such that $\sin \alpha \sin \delta = \sin \beta \sin \gamma$ and $\alpha + \beta = \gamma + \delta < 180^{\circ}$, then $\alpha = \gamma$ and $\beta = \delta$.

Denote by $M, N, R$ and $S$ the tangent points of the incircle of $ABCD$ centered at $O$ with the sides $AB, BC, CD$ and $DA$, respectively. Then the points $A, M, O, P$ and $S$ lie on the circle with diameter $AO$ and $\text{APM = AM 2 = AS 2 = APS}$.

Analogously, $\text{CPR = CPN}$ and hence $\text{SPR = MPN}$.

The Sine theorem for $\triangle BPM$ and $\triangle BPN$ gives
$$\text{MPB PMB = BM BP = BN BP = BPN BNP}$$
and therefore $\text{MPB NPB = PMB PNB}$.

Since
$$\text{PMB = 180 - AMP = 180 - AOP PNB = 180 - CNP = 180 - COP}$$
then $\text{MPB NPB = AOP COP}$.

We get in the same way that
$$\text{SPD RPD = AOP COP .}$$

Applying the fact mentioned above with $\text{= MPB}$, $\text{= NPB}$, $\text{= SPD}$ and $\text{= RPD}$ we conclude that $\text{MPB = SPD}$ and therefore $\text{APB = APM + MPB = APS + SPD = APD}$.