A question about the converse intercept theorem

Question

My question came from a task, which I have not solved yet. I tried to summarise my question beyond the scope of the specific problem, so let's look at the following diagram: We want to know if $p_1\parallel p_2$. Is it appropriate and sufficient $\dfrac{OP_1}{OP}=\dfrac{OQ_1}{OQ}$ for the lines to be parallel? How can I show this using vectors? Does the converse of the intercept theorem allow us to conclude they are parallel? (We have $\dfrac{OP_1}{OP}=\dfrac{OQ_1}{OQ}$. Does this mean $p_1 \parallel p_2$?)

Answer 1

Draw perpendiculars from $Q$ and $Q_1$ on line $PP_1$ to touch this line at points A and B respectively. RA triangles $OAQ_1$ and $OBQ$ are similar and we have:

$\frac{OQ_1}{OQ}=\frac{Q_1A}{QB}$

So triangles $P_1Q_1A$ and $PBQ$are similar and we can write:

$\frac{OB}{OA}=\frac{OP}{OP_1}$

$\frac{OA+AP_1}{OB}=\frac{OB+BP}{OB}$⇒$\frac{AP_1}{BP}=\frac{OA}{OB}=\frac{AQ_1}{BQ}$

That means RA triangles $AP_1Q_1$ and $BPQ$ are similar therefore angles $QPB$ and $Q_1P_!A$ are equal, which in turn means lines $P$ and $P_1$ are parallel.

So it seems reasonable to consider relation $\frac{OP_1}{OP}=\frac{OQ_1}{OQ}$ as necessary and sufficient condition for lines $P$ and $P_1$ to be parallel.

Answer 2

The statement ist false: You can consider a point $Q_1'\neq Q_1$ which is on the Line $OQ$ such that $\vert OQ_1 \vert= \vert OQ_1' \vert$, but $P_1Q_1'$ is not parallel to $PQ$.

So in general the intercept theorem is not reversable if you state it with lines. If you state it only for half-lines the converse is true.