Show that $|AB|\textbf{p}+|BC|\textbf{q}+|CA|\textbf{r}=\textbf{0}$ (without using the cross product). Let $ABC$ be a triangle, and p, q, r be the unit vectors perpendicular to
$AB$, $BC$, $CA$ directed inside the triangle. Show that $|AB|\textbf{p}+|BC|\textbf{q}+|CA|\textbf{r}=\textbf{0}$.
I tried using that the dot product of two perpendicular vectors is zero and that $\textbf{AB}+\textbf{BC}+\textbf{CA}=\textbf{0}$ but...

take $|AB|\textbf{p}=\vec{A'B'},|BC|q=\vec{B'C'},|BC|q=\vec{C'A1'}$, the 3 sides can form a $ \triangle A'B'C'  \cong \triangle ABC$ where $ A1'=A'$.then the answer is clear.you may need some explanation for the angles. 
