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 $\overrightarrow{BD}$ in terms of $\bar p$ and $\bar q$.

Two vectors acting simultaneously from a common point are added by the Parallelogram Law: if AB=a and AD=b are adjacent sides of a parallelogram, the diagonal AC through the common vertex is a+b. Equivalently, the Triangle Law adds vectors placed tip-to-tail: if AB=a and BC=b, then AC=a+b closes the triangle. Extending the triangle law around several vectors placed tip to tail gives the polygon law, and when the vectors return to the starting point their sum is zero -- which is why, for any triangle ABC, AB+BC+CA=0. Addition is commutative (a+b=b+a) and associative, has the zero vector as identity, and every vector has an additive inverse (its negative). Subtraction a-b is simply addition of the negative, a+(-b), carried out by reversing the arrow for b before applying the triangle law. The Triangle Inequality |a+b| <= |a|+|b| is an immediate consequence of the triangle law, since one side of a triangle can never exceed the sum of the other two.