A question on plane geometry $ABCD$ is a square with centre $O$. Let $P$ be the midpoint of $OB$ and $Q$ the midpoint of $CD$. Prove, using vectors and complex numbers, that $AP$ and $PQ$ are perpendicular segments of the same length.
If $AP$ and $PQ$ are perpendicular, which means their product equal to $0$, I tried many ways but can't get $(\vec{AP})(\vec{PQ})=0$ or $(P-Q)(A-P)=0$, and I let centre $O$ be the origin. Any one has hint?
 A: Let the square be centered at the origin and the length of its edge be $4x$. 
Name the corners putting $A$ on the lower-left corner and going counterclockwise. We have
$$ A=(-2x,\, -2x), \qquad P = (x,\,-x), \qquad AP = P-A = (3x,\,x) $$
and
$$ Q=(0,\, 2x), \qquad PQ = Q-P = (-x,\,3x).  $$
Hence it follows that
$$ PQ\cdot AP = -3x^2 + 3x^2 = 0 \quad \text{and} \quad |PQ|=|AP|=\sqrt{10}\cdot x $$

Of course you don't really need $x$. You can work under the assumption that the edge has length $4$, then the general case follows due to the symmetry of the problem.
A: Let the square be given by $A=-2i$, $B=2$, $C=2i$, $D=-2$ in the complex plane. Then $P=1$ and $Q=-1+i$. It follows that
$$Q-P=-2+i=i(1+2i)=i(P-A)\ ,$$
which proves the assertion, as multiplication by $i$ amounts to a rotation by $90^\circ$ in the counterclockwise direction.
A: Supposed it is known that the diagonals of a square bisect each other, are equal and perpendicular, let $\mathbf r$ denote the vector $\overrightarrow{OR}\,$, i.e. the position vector of the point $R$ with respect to the center of the given square.
Then $$\mathbf q-\mathbf p=\frac 12(\mathbf c+\mathbf d)-\frac 12\mathbf b=\frac 12(\mathbf c-\mathbf b)-\frac 12\mathbf b=\frac 12\mathbf c-\mathbf b$$$$\mathbf p- \mathbf a=\frac 12\mathbf b+\mathbf c$$hence $$(\mathbf q-\mathbf p)\cdot(\mathbf p- \mathbf a)=\frac 12(c^2-b^2)=0$$Moreover $$(\mathbf q-\mathbf p)^2=\frac 54b^2=(\mathbf p-\mathbf a)^2$$
