Step 1 - normalise the original vectors. So define $\vec{\dot{a}} = \frac{\vec{a}}{|\vec{a}|}$ and similarly for $\vec{\dot{b}}$, then let $\vec{\dot{c}} = \vec{\dot{a}} + \vec{\dot{b}}$. It should be pretty simple to prove that the direction of $\vec{\dot{c}}$ is the same as the one of $\vec{c}$ in your post.
Step 2 - Find the angle between the new proposed bisector and the original vectors. So define $\alpha$ as the angle between $\vec{a}$ and $\vec{c}$, and then $\vec{\dot{a}} \cdot \vec{\dot{c}} = |\vec{\dot{a}}||\vec{\dot{c}}|\cos{\alpha} = |\vec{\dot{c}}|\cos{\alpha}$ since we set $|\vec{\dot{a}}| = 1$ in the first step. Similarly if $\beta$ is the angle between $\vec{b}$ and $\vec{c}$, then $\vec{\dot{b}} \cdot \vec{\dot{c}} = |\vec{\dot{c}}|\cos{\beta}$.
But, from the way they've been defined, $\vec{\dot{a}} \cdot \vec{\dot{c}} = \vec{\dot{a}} \cdot \vec{\dot{a}} + \vec{\dot{a}} \cdot \vec{\dot{b}} = |\vec{\dot{a}}| + \vec{\dot{a}} \cdot \vec{\dot{b}} = 1 + \vec{\dot{a}} \cdot \vec{\dot{b}}$, and you can show that the other dot product has the same value. So you can conclude that $\cos{\alpha} = \cos{\beta}$, and then all you have to do is show that the angles are in the same quadrant, and hence must be equal.