Let N and M be midpoints of sides OA and AB of triangle OAB…

Let $N$ and $M$ be midpoints of sides $OA$ and $AB$ of triangle $OAB$. Express analytically and decompose geometrically: $1^\circ$ vectors $\overrightarrow{OM}$ and $\overrightarrow{MN}$ in terms of $\overrightarrow{OA}=\vec{a}$ and $\overrightarrow{OB}=\vec{b}$; $2^\circ$ vectors $\overrightarrow{ON}$ and $\overrightarrow{BN}$ in terms of $\vec{b}$ and $\overrightarrow{OM}=\vec{c}$.

Explanation of Method

Triangle midpoints — vector decomposition via two different bases.

Step 1: Setup

Let $\overrightarrow{OA} = \vec{a}$ and $\overrightarrow{OB} = \vec{b}$. Since $N$ is the midpoint of $OA$, $\overrightarrow{ON} = \frac{1}{2}\vec{a}$. and $M$ is the midpoint of $AB$, so $\overrightarrow{AM}=\tfrac12\overrightarrow{AB}$.

Step 2: Part $1^\circ$: $\overrightarrow{OM},\ \overrightarrow{MN}$ via $\vec{a},\ \vec{b}$

$\overrightarrow{OM}$

\[ \overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \overrightarrow{OA} + \frac{1}{2}\overrightarrow{AB}, \qquad \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \vec{b} - \vec{a} \]

\[ \overrightarrow{OM} = \vec{a} + \frac{1}{2}(\vec{b}-\vec{a}) = \frac{1}{2}\vec{a} + \frac{1}{2}\vec{b} \]

\[\overrightarrow{OM} = \frac{1}{2}(\vec{a}+\vec{b})\]

Geometric decomposition: $\overrightarrow{OM}$ is the diagonal of the parallelogram built on $\tfrac{1}{2}\vec{a}$ and $\tfrac{1}{2}\vec{b}$, i.e. the half-sum of $\vec{a}$ and $\vec{b}$.

$\overrightarrow{MN}$

\[ \overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} = \frac{1}{2}\vec{a} - \left(\frac{1}{2}\vec{a}+\frac{1}{2}\vec{b}\right) = -\frac{1}{2}\vec{b} \]

\[\overrightarrow{MN} = -\frac{1}{2}\vec{b}\]

Geometric decomposition: $\overrightarrow{MN}$ is antiparallel to $\vec{b}$ with half its magnitude — the midline theorem: $MN \parallel OB$ and $MN = \tfrac{1}{2}OB$.

$O$ $A$ $B$ $N$ $M$ $\vec{a}$ $\vec{b}$

$\vec{a}=\overrightarrow{OA}$
$\vec{b}=\overrightarrow{OB}$
$\overrightarrow{OM}=\tfrac12(\vec a+\vec b)$
$\overrightarrow{MN}=-\tfrac12\vec b$

Part $1^\circ$: $N$ is the midpoint of $OA$, $M$ the midpoint of $AB$; $MN\parallel OB$.

Step 3: Part $2^\circ$: $\overrightarrow{ON},\ \overrightarrow{BN}$ via $\vec{b},\ \vec{c}$

From Part $1^\circ$, $\vec{c} = \tfrac{1}{2}\vec{a} + \tfrac{1}{2}\vec{b}$. Solving for $\vec{a}$: \[\frac{1}{2}\vec{a} = \vec{c} - \frac{1}{2}\vec{b} \quad\Longrightarrow\quad \vec{a} = 2\vec{c} - \vec{b}.\]

$\overrightarrow{ON}$

\[ \overrightarrow{ON} = \frac{1}{2}\vec{a} = \frac{1}{2}(2\vec{c}-\vec{b}) = \vec{c} - \frac{1}{2}\vec{b} \]

\[\overrightarrow{ON} = \vec{c} - \frac{1}{2}\vec{b}\]

Geometric decomposition: start with $\overrightarrow{OM} = \vec{c}$ and add $\overrightarrow{MN}=-\tfrac{1}{2}\vec{b}$; $\overrightarrow{ON}$ is the diagonal of the parallelogram built on $\vec{c}$ and $-\tfrac{1}{2}\vec{b}$.

$\overrightarrow{BN}$

\[ \overrightarrow{BN} = \overrightarrow{ON} - \overrightarrow{OB} = \left(\vec{c} - \frac{1}{2}\vec{b}\right) - \vec{b} = \vec{c} - \frac{3}{2}\vec{b} \]

\[\overrightarrow{BN} = \vec{c} - \frac{3}{2}\vec{b}\]

Geometric decomposition: $\overrightarrow{BN}$ is the diagonal of the parallelogram built on $\vec{c}$ and $-\tfrac{3}{2}\vec{b}$.

$O$ $A$ $B$ $N$ $M$ $\vec{b}$

$\vec c=\overrightarrow{OM}$
$\overrightarrow{MN}=-\tfrac12\vec b$
$\overrightarrow{ON}=\vec c-\tfrac12\vec b$
$\overrightarrow{BN}=\vec c-\tfrac32\vec b$

Part $2^\circ$: $\overrightarrow{ON}=\overrightarrow{OM}+\overrightarrow{MN}$ and $\overrightarrow{BN}=\overrightarrow{BO}+\overrightarrow{ON}$.

Result

Vector	Expression	Basis
$\overrightarrow{OM}$	$\dfrac{1}{2}\vec{a}+\dfrac{1}{2}\vec{b}$	$\vec{a},\,\vec{b}$
$\overrightarrow{MN}$	$-\dfrac{1}{2}\vec{b}$	$\vec{a},\,\vec{b}$
$\overrightarrow{ON}$	$\vec{c}-\dfrac{1}{2}\vec{b}$	$\vec{b},\,\vec{c}$
$\overrightarrow{BN}$	$\vec{c}-\dfrac{3}{2}\vec{b}$	$\vec{b},\,\vec{c}$