In parallelogram $ABCD$ there is a point $P$ inside it There is a parallelogram $ABCD$ and a point $P$ inside it, where $\lvert CP\rvert=\lvert CB\rvert$. Is there a way to prove that a line connecting midpoint of $AP$ and midpoint of $DC$ is perpendicular to a line connecting points $B$ and $P$?
 
Unfortunately I can't post inline pictures yet.
Thanks
Greg
 A: This answer uses vectors.
Let $\vec{CB}=\vec b,\vec{CD}=\vec d$.
Then, we can write $\vec{CP}=m\vec b+n\vec d$ where $m,n\in\mathbb R$ and have $$|\vec{CB}|^2=|\vec{CP}|^2\iff |\vec b|^2=m^2|\vec b|^2+n^2|\vec d|^2+2mn\vec b\cdot\vec d\tag1$$
Now we have
$\vec{FG}=\vec{CG}-\vec{CF}=\frac 12(\vec{CP}+\vec{CA})-\frac 12\vec d=\frac 12(m\vec b+n\vec d+\vec b+\vec d)-\frac 12\vec d=\frac 12(m\vec b+n\vec d+\vec b)$
and
$\vec{BP}=\vec{CP}-\vec{CB}=m\vec b+n\vec d-\vec b$
So,
$$\begin{align}\vec{FG}\cdot\vec{BP}&=\frac 12(m\vec b+n\vec d+\vec b)\cdot (m\vec b+n\vec d-\vec b)\\\\&=\frac 12(m|\vec b|^2+n^2|\vec d|^2+2mn\vec b\cdot\vec d-|\vec b|^2)\end{align}$$
We see that this equals $0$ from $(1)$.
A: Let X be the midpoint of BP. Then, $\angle CXP = 90^0$ because $\triangle CBX \cong \triangle CPX$.

By midpoint theorem, GX is equal and parallel to (half AB).
But (half AB) is equal and parallel to (half DC) = FC. This means FGXC is a parallelogram. 
Then, $\angle GYX = \angle CXP = 90^0$
A: Let $A=(0,0),B=(a,0),C=(b,c),D=(b-a,c)$ the coordinates of the parallelogram.
►Circle $\Gamma$ centered at $(b,c)$ and radius $\sqrt{c^2+(b-a)^2}$ has equation
 $$x^2-2bxy^2-2cy=a^2-2ab$$ Let $P=(x,y)\in\Gamma$
►Line $FG$ and line $BP$ have pentes $\dfrac{y-2c}{x-2b+a}$ and $\dfrac{-x+a}{y}$
Is it $\dfrac{y-2c}{x-2b+a}$ equal to $-\dfrac{y}{x-a}$? (in other words is it $mm'+1=0$?) A simple verification shows that the answer is YES because
$$\dfrac{y-2c}{x-2b+a}=-\dfrac{y}{x-a}\iff x^2-2bxy^2-2cy=a^2-2ab$$
