In triangle ABC , side BC is divided by point D in the ratio…

In triangle $ABC$, side $BC$ is divided by point $D$ in the ratio $m:n$. Express vector $\overrightarrow{AD}$ in terms of vectors $\overrightarrow{AB}=\vec{c}$ and $\overrightarrow{AC}=\vec{b}$.

Step 1: Setup

Point $D$ lies on $BC$ with $\dfrac{BD}{DC} = \dfrac{m}{n}$, so $BD$ is $\dfrac{m}{m+n}$ of the whole side: \[\overrightarrow{BD} = \frac{m}{m+n}\,\overrightarrow{BC}, \qquad \overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AC} = \vec{b} - \vec{c}.\]

Step 2: Find $\overrightarrow{AD}$

Along the path $A \to B \to D$: \[\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BD} = \vec{c} + \frac{m}{m+n}(\vec{b}-\vec{c}) = \left(1 - \frac{m}{m+n}\right)\vec{c} + \frac{m}{m+n}\vec{b} = \frac{n}{m+n}\,\vec{c} + \frac{m}{m+n}\,\vec{b}\]

\[\overrightarrow{AD} = \frac{n\,\vec{c} + m\,\vec{b}}{m+n}\]

Step 3: Check via the path $A \to C \to D$

$\overrightarrow{CD}$ points from $C$ towards $B$ and has $\tfrac{n}{m+n}$ of the length of $CB$: $\overrightarrow{CD} = \frac{n}{m+n}\,\overrightarrow{CB} = \frac{n}{m+n}(\vec{c}-\vec{b})$.

\[ \overrightarrow{AD} = \vec{b} + \frac{n}{m+n}(\vec{c}-\vec{b}) = \frac{m}{m+n}\vec{b} + \frac{n}{m+n}\vec{c} \;\checkmark \]

Step 4: Special cases

Midpoint: $m = n$

\[ \overrightarrow{AD} = \frac{n\vec{c}+n\vec{b}}{2n} = \frac{\vec{b}+\vec{c}}{2} \]

This is the median formula: the median from $A$ is half the sum of the adjacent side vectors.

$D = B$: $m = 0$

\[ \overrightarrow{AD} = \frac{n\vec{c}}{n} = \vec{c} = \overrightarrow{AB} \;\checkmark \]

$D = C$: $n = 0$

\[ \overrightarrow{AD} = \frac{m\vec{b}}{m} = \vec{b} = \overrightarrow{AC} \;\checkmark \]