Lemma: Let $WXYZ$ be an arbitrary quadrilateral, and let $M$ and $N$ be the midpoints of $WZ$ and $XY$, respectively. Then $2MN \leq WX + YZ$. Proof: Consider the points as vectors. We have \[ 2|M-N| = 2 \left |\frac{W+Z}{2} - \frac{X+Y}{2} \right | = |(W-X) + (Z-Y)| \leq |W-X| + |Z-Y|, \] that is, $2MN \leq WX + YZ$, as desired. $\blacksquare$ Let $\omega_1$ be the circumcircle of $\triangle APD$ and let $\omega_2$ be the circumcircle of $\triangle BCP$. Let $E$ be the intersection of lines $AB$ and $CD$ (possibly at infinity), let $\theta = \angle AED$ (and set $\theta = 0$ if $E = \infty$), and suppose without loss of generality that $A$ lies between $E$ and $B$. Then we have \[ \angle DPA = 180^{\circ} - (\angle PAD + \angle ADP) = (\angle BAD + \angle ADC - \theta) - (\angle PAD + \angle ADP) = \angle BAP + \angle PDC - \theta \leq 90 - \theta, \] and \[ \angle CPB = 180^{\circ} - (\angle PBC + \angle BCP) = (\angle ABC + \angle BCD + \theta) - (\angle PBC + \angle BCP) = \angle ABP + \angle PBC + \theta \leq 90 + \theta. \] Now consider the circles $\Gamma_1$ and $\Gamma_2$ such that the directed angle $\angle DXA$ equals $90^{\circ} - \theta$ for all $X \in \Gamma_1$, and the directed angle $\angle CXA$ equals $90^{\circ} + \theta$ for all $X \in \Gamma_2$. We claim that $\Gamma_1$ and $\Gamma_2$ cannot intersect twice. Suppose for the sake of contradiction there exists some point $P'$ lying on both $\Gamma_1$ and $\Gamma_2$. Since $\angle DP'A \geq \angle DPA$, $P'$must lie within $\omega_1$, and since $\angle CP'B \geq \angle CPB$, $P'$ lies within $\omega_2$. But since $\omega_1$ and $\omega_2$ are tangent at $P$, we must have $P' = P$, so our claim is proven. We now have two cases: Case 1: $E = \infty$. We have $\angle DPA = \angle CPB =90^{\circ}$. Let $M$ and $N$ be the midpoints of $AD$ and $BC$, respectively. Because $\Gamma_1$ and $\Gamma_2$ have diameters of $AD$ and $BC$, they are centered at $M$ and $N$, respectively, and have radii of $\frac{AD}{2}$ and $\frac{BC}{2}$, respectively. Since $\Gamma_1$ and $\Gamma_2$ do not intersect intersect, we have $MN \geq \frac{AD+BC}{2}$, (since $\frac{AD+BC}{2}$ is the sum of the radii of $\Gamma_1$ and the $\Gamma_2$.) But since $AB || CD$, we have $MN = \frac{AB+CD}{2}$, so $AB + CD \geq AD + CB$, as desired. Case 2: $E \neq \infty$. Consider an inversion about $E$ with radius 1, and suppose it maps $A,B,C,D,\Gamma_1, \Gamma_2$ to $A', B', C', D', \Gamma_1', \Gamma_2'$, respectively. We have that $B'$ is between $E$ and $A'$, $C'$ is between $D'$ and $E'$, the interior of $ABCD$ is mapped to the interior of $A'B'C'D'$, and that $\Gamma_1'$ and $\Gamma_2'$ intersect in at most one point. Furthermore, for any point $X \in \Gamma_1$ in the interior of $ABCD$, we have \[ 90^{\circ} - \theta = \angle DXA = \angle DXE + \angle EXA = \angle ED'X' + \angle EA'X' = 360^{\circ} - \theta - \angle D'XA' = \angle D'XA' - \theta, \] so $\angle D'XA' = 90^{\circ}$, whence $A'D'$ is a diameter of $\Gamma_1'$. Similarly, we can find that $B'C'$ is a diameter of $\Gamma_2'$. Let $M$ and $N$ be the midpoints of $A'D'$ and $B'C'$, respectively. Because $\Gamma_1'$ and $\Gamma_2'$ have $A'D'$ and $B'C'$ as their diameters, respectively, they are centered at $M$ and $N$, respectively, and have radii of $\frac{A'D'}{2}$ and $\frac{B'C'}{2}$, respectively. Because they do not intersect, $MN \geq \frac{A'D' + B'C'}{2}$. By the lemma, $2MN \leq A'B' + C'D'$, so we have $A'D' + B'C' \leq A'B' + C'D'$. We wished to show that $AB + CD \geq AD + BC$. This is equivalent to $EB- EA + EC - ED - AD - BC \geq 0$ which, after our inversion, is equivalent to \[ \frac{1}{EB'} - \frac{1}{EA'} + \frac{1}{EC'} - \frac{1}{ED'} - \frac{A'D'}{EA' \cdot ED'} - \frac{B'C'}{EB' \cdot EC'} \geq 0, \] or \begin{align*} \frac{EB' + EC' - B'C'}{EB' EC'} \geq \frac{EA' + ED' + A'D'}{EA' \cdot ED'} \tag{2}. \end{align*} Now, by the law of cosines, we have \begin{align*} 1 + \cos \angle E &= \frac{EB'^2 + EC'^2 - B'C'^2}{2EB' \cdot EC'} + 1 = \frac{(EB' + EC')^2 - B'C'^2}{2EB' \cdot EC'} \\ & = \frac{(EB' + EC' - B'C')(EB' + EC' + B'C')}{2EB' EC'}. \end{align*} Similarly, \[ 1 + \cos \angle E = \frac{(EA' + ED' + A'D')(EA' + ED' - A'D')}{2EA' \cdot E'D'}. \] Thus, if we divide both sides of (2) by $2(1 + \cos \angle E)$, we see that (2) is equivalent to \[ \frac{1}{EB' + EC' + B'C'} \geq \frac{1}{EA' + ED' - AD'} \iff EA' + ED' \geq EB' + EC' + BC' + AD' . \] Recalling that $EA' = EB' + B'A'$ and $ED' = EC' + C'D'$, we see that this last inequality is equivalent to $A'B' + C'D' \geq A'D' + B'C'$, which we have proven.

Conisder the second intersection $Q$ of circumcircles $APB$ and $CPD$. First, we show that the circumcircles $AQD$ and $BQC$ are tangent as well. It suffices to show that $\angle DAQ + \angle CBQ = \angle CQD$, since then the tangents at $Q$ to the two circumcircles $AQD$ and $BQC$ must coincide. But since $\angle DAQ + \angle CBQ = \angle DAP - \angle PAQ + \angle CBP + \angle PBQ$ and $\angle PAQ = \angle PBQ$, we have $\angle DAQ + \angle CBQ = \angle DAP + \angle CBP = \angle CPD = \angle CQD$ as desired. Thus the circumcircles $AQD$ and $BQC$ are tangent. Next, note that $\angle BQC = \angle BQP + \angle CQP = \angle BAP + \angle CDP \leq 90^{\circ}$ and similarly $\angle AQD \leq 90^{\circ}$. It follows that there exists at most one point $R$ with $\angle ARD = \angle BRC = 90^{\circ}$. Therefore, the distance from the centers of the two circles with diameters $AD$ and $BC$ must be at least as great as the sum of their radii; i.e. $MN \geq \frac{1}{2} (AD + BC)$ if $M$ and $N$ are the midpoints of $AD$ and $BC$, respectively. Finally, since $AB + CD \geq 2MN$ is equivalent to the triangle inequality in complex numbers, it follows that $AB + CD \geq AD + BC$.