OABC is a parallelogram and M is the midpoint of BC

Key concept: This is a vector ratio problem inside a parallelogram. Choose a convenient vector basis from $\overrightarrow{OA}$ and $\overrightarrow{OC}$, then express all other points in terms of those vectors.

Step by step

Let

$\overrightarrow{OA}=\mathbf{a},\qquad \overrightarrow{OC}=\mathbf{c}$

Since $OABC$ is a parallelogram,

$\overrightarrow{OB}=\mathbf{a}+\mathbf{c}$

Also, $M$ is the midpoint of $BC$. Now

$\overrightarrow{OB}=\mathbf{a}+\mathbf{c},\qquad \overrightarrow{OC}=\mathbf{c}$

so the midpoint $M$ has position vector

$\overrightarrow{OM}=\frac{\overrightarrow{OB}+\overrightarrow{OC}}{2}=\frac{(\mathbf{a}+\mathbf{c})+\mathbf{c}}{2}=\frac{\mathbf{a}+2\mathbf{c}}{2}$

Let $Q$ be the intersection of $AM$ and $OB$.

Along line $AM$

Since $AQ=hAM$,

$\overrightarrow{OQ}=\overrightarrow{OA}+h\,\overrightarrow{AM}$

Now

$\overrightarrow{AM}=\overrightarrow{OM}-\overrightarrow{OA}=\frac{\mathbf{a}+2\mathbf{c}}{2}-\mathbf{a}=\frac{-\mathbf{a}+2\mathbf{c}}{2}$

Hence

$\overrightarrow{OQ}=\mathbf{a}+h\left(\frac{-\mathbf{a}+2\mathbf{c}}{2}\right)$

$\overrightarrow{OQ}=\left(1-\frac h2\right)\mathbf{a}+h\mathbf{c}$

Along diagonal $OB$

Since $OQ=k\,OB$ and $\overrightarrow{OB}=\mathbf{a}+\mathbf{c}$,

$\overrightarrow{OQ}=k(\mathbf{a}+\mathbf{c})=k\mathbf{a}+k\mathbf{c}$

Compare coefficients

Equate the two expressions for $\overrightarrow{OQ}$:

$\left(1-\frac h2\right)\mathbf{a}+h\mathbf{c}=k\mathbf{a}+k\mathbf{c}$

So

$1-\frac h2=k$

and

$h=k$

Substitute $h=k$ into the first equation:

$1-\frac h2=h$

$1=\frac{3h}{2}$

$h=\frac{2}{3}$

Therefore

$k=\frac{2}{3}$

Answer

$\boxed{h=\frac{2}{3},\quad k=\frac{2}{3}}$

Pitfall alert

Do not assume the midpoint of $BC$ has vector halfway between $O$ and $C$. The midpoint must be taken between the endpoints $B$ and $C$. Also, remember that $AQ=hAM$ means $Q$ divides $AM$ from $A$ toward $M$.

Try different conditions

A coordinate method gives the same result. If $O=(0,0)$, $A=(1,0)$, and $C=(0,1)$, then $B=(1,1)$ and $M=(\tfrac12,1)$. Solving the line intersection of $AM$ and $OB$ again gives $Q$ at two-thirds of the way along both segments.

Further reading

parallelogram vector, midpoint, section ratio