Let $R=AC\cap BD$; $BR$ is symmedian of $\Delta ABC$, so $\frac{AR}{CR}=\left(\frac{AB}{BC}\right)^2$, but $\frac{AR}{CR}=\frac{AB}{CD}$, so $BC^2=AB\cdot CD$, consequently $\frac{BR}{RD}=\left(\frac{BC}{CD}\right)^2\implies CR$ is symmedian of $\Delta BCD$. Best regards, sunken rock

Let $M$ be the intersection of $AC$ and $BD$ and $R$ be the intersection of $CQ$ and $BR$ (these possibly extended). Since $P$ and $Q$ are midpoint of $AC$ and $BD$, we have that $PQ$ bisects $BC$. Hence, By Ceva's theorem in $\triangle PBC$ with point $Q$ we have $\frac{PR}{RB}=\frac{PM}{MC}$ and hence $MR\parallel CB$. Now $\angle QPR=\angle ABP=\angle CBD=\angle CBM=\angle RMB$ (since $AB\parallel PQ$, by the given condition and since $BC\parallel RM$ respectively). Thus $PRQM$ is a cyclic quadrilateral. Hence $\angle BCQ=\angle MQR=\angle APR$. In $\triangle APB$ and $\triangle CQB$ we now have two pairs of equal angles, hence the third must be equal as well: $\angle BCQ=\angle PAB=\angle CAB=\angle ACD$ (since $AB\parallel CD$) as required.

Let $BP$ cuts $CD$ at $E$, then $ABCE$ is a parallelogram and $\angle EAB = \angle DCB$, thus $\Delta EAB\sim \Delta DCB$, as $P$ is the midpoint of $EB$ and $Q$ the midpoint of $DB$, $\angle APB = \angle CQB$, thus $\angle BCQ = \angle PAB = \angle ACD$

compare the problems http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2254841&sid=c8d0d7fa7996a8b8d03385a7f11e1e3f#p2254841 you can easily see that USA problem is more harder and more beautiful.

This is not hard. Let $M=AC\cap BD$, and let $X,Y,Z$ be the feet of the altitudes from $M$ to $AB,BC,CD$ respectively. Note that by the angle equality we have that $BM$ and $BP$ are isogonal conjugates of each other, but since $AP$ is a median of $\triangle ABC$ it follows that $BM$ is the $B$-symmedian of said triangle. Therefore $\dfrac{MY}{MX}=\dfrac{BM}{AB}$. But since $\triangle ABM\sim\triangle CDM$, $\dfrac{MX}{MZ}=\dfrac{AB}{CD}$, and multiplying these two ratios together gives $\dfrac{MY}{MZ}=\dfrac{BC}{CD}$. Therefore $CQ$ is the $C$-symmedian of $\triangle BCD\implies $ $CM$ and $CQ$ are isogonal conjugates $\implies \angle BCQ=\angle ACD$ as desired. $\blacksquare$

Let the circumcircle of triangle ABC intersects BD at E. Here AE is a symmedian of ABC. So ABCD is a harmonic quad. So AC will be a symmedian of triangle BCE.

Seems too easy for PAMO. Proof: Let $O=^{Def} AC \cap BD$. The angle condition tells us $\angle ABP =$$ \angle CBD \implies \angle ABP = \angle CBO \iff \angle ABO = \angle CBP$. Since $P$ is the midpoint of $AC$, it implies that $BP$ is the median of $\triangle ABC$ and so $BO$ is the $B-$symmedian of the triangle. So the problem is equivalent to proving that $CO$ is the $C-$symmedian of $\triangle BCD$. $\rightarrow $ From the fact that $BO$ is the symmedian, we have : $\dfrac { AO}{CO}=\dfrac {AB^2} {BC^2}$. Also since $AB \parallel CD \implies \triangle AOB \sim \triangle COD \implies \dfrac {AO}{CO} = \dfrac{AB}{CD} \iff BC^2 = AB.CD$. Hence, from the triangle similarity, we obtain $\dfrac{BO}{DO} = \dfrac{AB}{CD} = \dfrac {AB .CD}{CD^2} = \dfrac{BC^2}{CD^2} \iff CO$ is the $C-$symmedian of $\triangle BCD$$\blacksquare$