Q.Let $A,B,C$ be the vertices of a triangle. Let $D,E,F$ be the midpoints of the sides $BC,CA,AB$ respectively. Show that $\vec{AD}+\vec{BE}+\vec{CF}=\vec 0$.

Addition — two equivalent pictures.

• Triangle law: if $\vec a=\vec{A_1B_1}$ and $\vec b=\vec{B_1B_2}$ (the tail of $\vec b$ placed at the tip of $\vec a$), then $\vec a+\vec b$ is the third side $\vec{A_1B_2}$, taken from the start of $\vec a$ to the end of $\vec b$. In words: if two vectors are represented, in magnitude and direction, by two sides of a triangle taken in order, their sum is the third side taken in the reverse order.
• Parallelogram law: if $\vec a=\vec{OA}$ and $\vec b=\vec{OB}$ share the initial point $O$, complete the parallelogram $OACB$; the diagonal $\vec{OC}$ through $O$ is $\vec a+\vec b$.

Both laws describe the same sum — the triangle law is just the parallelogram law applied to half of the parallelogram.

Key results proved from the triangle law:

• If $\vec a,\vec b,\vec c$ are the three sides of a triangle taken in order (tip to tail, returning to the start), $\vec a+\vec b+\vec c=\vec 0$.
• Vector addition is associative: $(\vec a+\vec b)+\vec c=\vec a+(\vec b+\vec c)$.
• $\vec a+\vec 0=\vec 0+\vec a=\vec a$ for every $\vec a$.
• $\vec a+(-\vec a)=\vec 0$, where $-\vec a$ (the reverse of $\vec a$) has the same magnitude as $\vec a$ but the opposite direction; if $\vec a=\vec{AB}$ then $-\vec a=\vec{BA}$.
• Vector addition is commutative: $\vec a+\vec b=\vec b+\vec a$ (proved by the parallelogram, since both diagonals lead to the same point $C$).
• Polygon law: for any chain of vectors placed tip to tail, $\vec{OA}+\vec{AB}+\vec{BC}+\vec{CD}+\vec{DE}=\vec{OE}$ — the sum is the single vector from the very first tail to the very last tip. …