proof of parallelogram using vector and midpoint

Question

$OPQR$ is a parallelogram. $T$ is the midpoint of $OR$. Show that $QT$ cuts the diagonal $PR$ in the ratio $2:1$.

Answer 1

Let $M$ be the intersection point of $QT$ and $PR$.

Since $M$ is on the $QT$, there exists $k\in\mathbb R$ such that $$\vec{QM}=k\vec{QT}=k\left(\vec{QR}+\frac 12\vec{QP}\right)=k\vec{QR}+\frac{k}{2}\vec{QP}\tag1$$ On the other hand, since $M$ is on the $PR$, there exists $l$ such that $$\vec{QM}=(1-l)\vec{QR}+l\vec{QP}\tag2$$

Since we have $(1)$ and $(2)$, we have $$k=1-l\ \text{and}\ \frac k2=l\Rightarrow (k,l)=\left(\frac 23,\frac 13\right).$$

So, the answer is $PM:MR=1-l:l=2/3:1/3=2:1$.

Answer 2

Let $S$ be the symmetric of $Q$ wrt $R$. Since $\frac{PQ}{TR}=2$, $S$ is the simmetric of $P$ wrt $T$, too. This gives that the intersection of $PR$ and $QT$ is the centroid of the triangle $PQS$, and since the medians cut themselves in thirds, the conclusion.