Geometry problem using vectors

Question

$P, Q, R$ are points on the sides $AB, BC$ and $CA$ respectively of triangle $ABC$ such that $AP:PB=BQ:QC=AR:RC=1:2$. Show that $PBQR$ is a parallelogram.

I am looking to solve this question by using vectors. Could someone share the required approach?

Answer 1

Hint: $\overrightarrow{PR} = \overrightarrow{AR} - \overrightarrow{AP} = \frac{1}{3}\,\overrightarrow{AC} - \frac{1}{3}\,\overrightarrow{AB} = \frac{1}{3}(\overrightarrow{AC} - \overrightarrow{AB}) = \frac{1}{3}\,\overrightarrow{BC} = \cdots\;$.

Answer 2

For Thales it is clear that $$\frac{CR}{RA}=\frac{CQ}{QB}=2\Rightarrow RQ//AB$$ Furthermore $$\frac{AP}{PB}=\frac{AR}{RC}=\frac 12\Rightarrow RP//QB$$ This is enough to be sure that $PBQR$ is a parallelogram. This is translate easily to lenguage of vectors.