Q.ABCD is a trapezium with AB parallel to DC and DC $=3AB$. M is the mid-point of DC, $\overrightarrow{AB}=\bar p$ and $\overrightarrow{BC}=\bar q$. Find $\overrightarrow{AM}$ in terms of $\bar p$ and $\bar q$.

Take $A$ as origin, so $A=\bar 0$, $B=\bar p$ (since $\overrightarrow{AB}=\bar p$). Since

$\overrightarrow{BC}=\bar q$, $C=\bar p+\bar q$. Since $DC\parallel AB$ with $DC=3AB$ and $ABCD$ is a

simple (non-self-intersecting) trapezium traversed in order, side $CD$ (from $C$ to $D$, continuing the

boundary) points opposite to $AB$: $\overrightarrow{CD}=-3\bar p$, so $D=C-3\bar p=\bar q+\bar p-3\bar p=\bar q-2\bar p$.

$M$ is the midpoint of $DC$: $M=\dfrac{D+C}{2}=\dfrac{(\bar q-2\bar p)+(\bar p+\bar q)}{2}=\dfrac{2\bar q-\bar p}{2}=\bar q-\dfrac{\bar p}2$.

Set A as the origin, build B and C using the given side vectors p and q, use the parallel-sides condition (with DC three times AB and pointing the opposite way around the boundary) to locate D, then find M as the midpoint of DC and subtract A's position vector (zero).

Assuming DC points the SAME way as AB (rather than opposite, as required for a simple non-self-crossing trapezium traversed in the stated vertex order), which flips the sign of the p-term throughout.