prove the angle of sum of vectors is always in the middle For two planar vectors $v_1, v_2$, and they are in the same quadrant.
Define $\angle(v_1)$ as the angle between $v_1$ and positive x-axis. And $\angle{v_1} \lt \angle(v_2)$.
Their sum $v_3 = v_1+v_2$, would this be true $\angle(v_1) \lt \angle(v_3) \lt \angle(v_2)$? can you prove it?
Here is what I have tried:
It should be easy to see $v_3$ is in the same quadrant with $v_1, v_2$, since $v_3= (a_3, b_3)=(a_1+a_2, b_1+b_2)$, which $\forall a_i$ have the same sign and $\forall b_i$ have the same sign.
Since all $v_1, v_2, v_3$ are in the same quadrant, the angle between any two of them is less than 90. The three of them would form an acute triangle.

Hint (elaborating on Tassle's comment):
You have that
$$\tan\angle v_1=\frac{a_1}{b_1},\ \tan\angle v_2=\frac{a_2}{b_2},\ \tan\angle v_3=\frac{a_1+a_2}{b_1+b_2}.$$
Since the $\tan$ function is monotonically increasing, it suffices to show that
$$\frac{a_1}{b_1}<\frac{a_1+a_2}{b_1+b_2}<\frac{a_2}{b_2}.$$
Since we know that $a_1/b_1<a_2/b_2$ (because $\angle v_1<\angle v_2$), can you prove the above inequality?
