$OPQR$ is a parallelogram, with $O$ the origin. $M$ is the midpoint of $PQ$. $OM$ and $RQ$ are extended to meet as $S$. $\vec{OP}=\mathbf{p}$ and $\vec{OR}=\mathbf{r}$.
In this question I have to find the position vector of $S$. I know that $\displaystyle \vec {OM}=\mathbf{p}+\frac{\mathbf r}{2}$. First I considered Pythagorean theorem, but it turned out that there is no right angle here hence I got bewildered.
On
You are almost there. Note that $\,s = \mu m = \mu (p + r/2)\,$ for some $\mu \in \mathbb{R}$ because $S$ lies on $OM$, and $\,s = \lambda p +r \,$ for some $\lambda \in \mathbb{R}$ because $S$ lies on $RQ$. Since $p, r$ are linearly independent:
$$ \lambda p + r = \mu (p + r/2) \quad\iff\quad \lambda = \mu\,, \;\; 1 = \mu / 2 $$
On
Here is a way using ratios on an $X$-figure:
$$\frac{|PM|}{|OP|}= \frac{|MQ|}{|QS|}\stackrel{|PM|=|MQ|=\frac{1}{2}|PQ|}{\Rightarrow} |OP| = |QS|$$
As $\vec{OP}$ and $\vec{QS}$ lie on parallel sides in a parallelogram and have the same direction you get $$\vec{OP} = \vec{QS} = p$$
Obviously we have also $$\vec{PM} = \vec{MQ} =\frac{1}{2}r$$ So, $$\vec{OS} = p+\frac{1}{2}r+\frac{1}{2}r+p = 2p+r$$
Forget the vectors and use the geometry that you might have already learned: $\triangle{OPM}$ and $\triangle{SQM}$ are clearly similar. We’re given that $QM=PM$, therefore they’re congruent, and so $SQ=OP$. Now use the parallelogram rule for vector addition to determine the position vector for $Q$, from which you can get to $S$ using the above information.