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

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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.

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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 $PM:MR=1-l:l=2/3:1/3=2:1$.

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  • $\begingroup$ pls solve this using "vector geometry" to show that QT cuts the diagonal PR in the ratio 2:1 $\endgroup$ – sekling Aug 8 '14 at 15:41
  • $\begingroup$ @sekling: I added it. Is there anything unclear? $\endgroup$ – mathlove Aug 8 '14 at 16:18
  • $\begingroup$ Pls show detail QM =(1−l)QR+lQP $\endgroup$ – sekling Aug 8 '14 at 16:20
  • $\begingroup$ @sekling: Since $M$ is on the line segment $PR$, there exists $l$ such that $\vec{QM}=(1-l)\vec{QR}+l\vec{QP}$. This means that $PM:MR=1-l:l$. (See the triangle $QPR$.) I'm sure that you can find this in your textbook. $\endgroup$ – mathlove Aug 8 '14 at 16:23
  • $\begingroup$ @sekling: We can write $\vec{QZ}=m\vec{QR}+l\vec{QP}$ for any point $Z$ on the plane. If $Z$ is a point on the line $PR$, then $l+m=1\iff m=1-l$. So we have $\vec{QZ}=(1-l)\vec{QR}+l\vec{QP}$. $\endgroup$ – mathlove Aug 8 '14 at 16:31

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