parallelogram and cyclic quadrilateral There is parallelogram $ABCD$ we choose points $P,Q,R$ which lies respectively at $AB,AC,AD$ such that $APQR$ is cyclic quadrilateral. Show that $AP\cdot AB+AR\cdot AD=AQ\cdot AC$
so far I haven't figured it out, any hints ? 

 A: I found there is a fairly simple vector proof of this. Let $A=0$ be the origin, and $B=v,\ D=w$ so that $C=v+w.$ Besides these vectors $v,w$ I'll use the vector $z$ to denote the center of the circle through $A,P,Q,R,$ and also $vw$ will be the dot product $v \cdot w$ and $v^2=v\cdot v$ for (any) vectors $v,w.$ Now let $P=av,\ R=bw,\ Q=c(v+w),$ where the scalars $a,b$ are in $(0,1)$ to reflect that $P,Q$ lie on sides $AB,AD$ respectively.
Note now that $AP\ AB=av^2$ and $AR\ AD = bw^2,$ as well as $AQ\ AC=c(v+w)^2.$ So in this vector notation the goal is to show that $av^2+bw^2=c(v+w)^2.$ Now we look at the center $z$ of the circle through $A,P,R$ (without at first assuming it goes through $Q$). This $z$ lies on the perpendicular bisector of segment $AP$, which gives $(z-(1/2)av)v=0$, and similarly $z$ lies on the perpendicular bisector of $AR$ so that also $(z-(1/2)bw)w=0.$ Even without locating $z$, we therefore know it has two properties:
$$zv=(1/2)av^2,\\ zw=(1/2)bw^2. \tag{1}$$
Now the circle goes through the origin also, so that its squared radius is the dot product $z^2.$ We can check that e.g. this circle indeed goes through $P=av$:
$$(z-av)^2=z^2-2a(zv)+a^2v^2=z^2-2a((1/2)av^2)+a^2v^2=z^2,$$
where we have used the substitution from $(1).$ Similarly this circle goes through $R=bw.$
Now the point $Q=c(v+w)$ lies on this same circle iff the squared distance from $Q$ to $z$ is $z^2.$ We have
$$(z-c(v+w))^2=z^2-2cz(v+w)+c^2(v+w)^2,$$
and this is $z^2$ iff $2z(v+w)=c(v+w)^2,$ which using $(1)$ to replace $zv,zw$ amounts to $av^2+bw^2=c(v+w)^2$ as required.
