[asy][asy] size(200); defaultpen(fontsize(8pt)); pair A = (0,0); pair B = (1,2); pair C = (4,2); pair D = (3,0); pair P = rotate(15,C)*B; pair M = midpoint(A--P); pair N = midpoint(C--D); pair Pp = extension(B,P,M,N); pair I = midpoint(P--B); draw(A--B--C--D--cycle^^C--P, lightblue); draw(M--N, orange); draw(B--P, magenta); draw(P--Pp, dashed+magenta); draw(A--P, palered); draw(M--I--C, dashed+purple); dot("$A$", A, dir(-100)); dot("$B$", B, dir(120)); dot("$C$", C, dir(40)); dot("$D$", D, dir(-30)); dot("$P$", P, dir(135)); dot("$M$", M, dir(140)); dot("$N$", N, dir(N)); dot("$I$", I, dir(I)); [/asy][/asy] Let $M$ be the midpoint of $AP$, $I$ be the midpoint of $BP$ and $N$ be the midpoint of $CD$. We will prove that $MICN$ is a parallelogram. We have $MI\parallel AB\parallel CD$ and $MI=\dfrac 12 AB$ since $MI$ is the midline of $\triangle PAB$. We also have $CN=\dfrac 12 CD=\dfrac 12 AB=MI$. Thus, $MICN$ is a parallelogram. From $MICN$ is a parallelogram we have $MN\parallel CI$. But $CI\perp BP$, as $CBP$ is an isoceles triangle, we have $MN\perp BP$.

Let $C = 0$, $B = 1$, $D = d$, $P = p$. Then $A = d+1$, $|p| = 1$, $m(AP) = \frac{d+p+1}{2}$, $m(CD)=\frac{d}{2}$. $-\frac{p-1}{\overline{p}-1} = \frac{p-1}{1-\frac{1}{p}} = p$ $\frac{\frac{d+p+1}{2}-\frac{d}{2}}{\frac{\overline{d}+\overline{p}+1}{2}-\frac{\overline{d}}{2}} = \frac{p+1}{\overline{p}+1} = \frac{p+1}{\frac{1}{p}+1} = p$ Thus $-\frac{p-1}{\overline{p}-1} = \frac{\frac{d+p+1}{2}-\frac{d}{2}}{\frac{\overline{d}+\overline{p}+1}{2}-\frac{\overline{d}}{2}}$. The proof is completed. $\blacksquare$