Show that PQRS is a parallelogram using vector method. Not using product of vectors.

Question

Let $\vec{a}$, $\vec{b}$, $\vec{c}$, $\vec{d}$ be the position vectors of the four distinct points P,Q,R,S respectively. If $\vec{b} - \vec{a}$ = $\vec{c} - \vec{d}$. Show that PQRS is a parallelogram.

Answer 1

For simplicity, I will use $P,Q,R,S$ for denoting both the points $P,Q,R,S$ and the vectors $\vec{OP},\vec{OQ},\vec{OR},\vec{OS}$ where $O$ is the origin. With this (useful) abuse of notation, the vector $\vec{PQ}$, for instance, is simply denoted through $Q-P$. In our problem we know that

$$ Q-P = R-S \tag{1}$$ hence $Q-P$ is parallel to $R-S$ and has the same length.
By rearranging both sides of $(1)$ we also get $$ Q-R = P-S \tag{2}$$ hence $Q-R$ is parallel to $P-S$ and has the same length.
So, by $(1)$ and $(2)$ it follows that the segments $QP,PS,SR,RQ$ are the sides of a parallelogram.

You may also use my suggestion in the comments: in the plane, $A,B,C,D$ are the vertices of a parallelogram iff the segments $AC$ and $BD$ share the same midpoint, that is the same as requesting that $A+C=B+D$ holds. In your case, $$ Q+S = R+P \tag{3} $$ is a consequence of $(1)$, too.