Label the four vertices of a parallelogram in counterclockwise order as OPQR. Prove that the line segment from $O$ to the midpoint of $PQ$ intersects the diagonal $PR$ in a point $X$ that is $1/3$ of the way from $P$ to $R$. I assumed $X$ is placed such that $\overrightarrow {PX}= a\ \overrightarrow {PR}$ and $\overrightarrow {XR}= b\ \overrightarrow {PR}$ but all I get after a bunch $f$ simplification is $a+b=1$ how do I proceed with this?
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Hint: $,X,$ is the centroid of $,\triangle OPQ,$ (why?). – dxiv May 07 '23 at 19:37
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The conclusion is immediately obvious with similarity. You can just "convert" that solution by using the usual vector notation – 冥王 Hades May 07 '23 at 20:16
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4Does this answer your question? Parallelogram geometry proof - found through a link of A parallelogram and a line joining a vertex to the midpoint of opposite side in the RHS "Related" list, which the proposed duplicate was closed as being a duplicate of it. However, the other duplicate listed, i.e., Proving two lines trisects a line, is more closely related to your problem. – John Omielan May 07 '23 at 22:50
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thanks a lot, @dxiv – Orpheus May 08 '23 at 14:30