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...
 A: 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. 
A: Given a vector $\vec a=(x,y)$ of $\mathbf R^2$ and a unit vector $\vec u \in \mathbf R^2$ perpendicular to $\vec a,$ the vector $|\vec a| \vec u$ has, as it is easy to see, the coordinates
$$
(-y,x), \text{ or } (y,-x).
$$ 
The condition on the choice of directions of the vectors $\vec p,\vec q,\vec r$ from your problem suggests, say, the first choice (we may interpret the condition as the requirement that the angle between $\vec a$ and $|\vec a|\vec u$ is $+90^\circ,$ in the positive, counterclockwise direction). Thus as the sum of three vectors
$$
\overrightarrow{AB}=(x_1,y_1),\overrightarrow{BC}=(x_2,y_2), \overrightarrow{CA}=(x_3,y_3)
$$
is equal to the zero vector, the sum of the vectors
$$
|AB|\vec p+|BC| \vec q+|CA|\vec r=(-y_1,x_1)+(-y_2,x_2)+(-y_3,x_3)
$$
is also equal to the zero vector. 
A: Let $\mathbf{n}=\vec{BC}\times \vec{AB}$ be a vector normal to the plane of the triangle, and let $\mathbf{\hat n}$ be the unit vector in the same direction.
Now, as the triangle is closed, we have $\vec{AB}+\vec{BC}+\vec{CA}=\mathbf{0}$. And we have that
$$
\mathbf{p}=\frac{\vec{AB}\times \mathbf{\hat n}}{|AB|}\\
\mathbf{q}=\frac{\vec{BC}\times \mathbf{\hat n}}{|BC|}\\
\mathbf{r}=\frac{\vec{CA}\times \mathbf{\hat n}}{|CA|}
$$
So, taking the cross product of our triangle vectors with the normal vector, we have
$$
|AB|\mathbf{p}+|BC|\mathbf{q}+|CA|\mathbf{r}=\mathbf{0}
$$
