Proof that$$\dfrac{\|\vec{CA}\|}{\|\vec{AX}\|} = \dfrac{\|\vec{CB}\|}{\|\vec{BX}\|}$$
having $$\dfrac{\vec{CA}}{\|\vec{CA}\|} + \dfrac{\vec{CB}}{\|\vec{CB}\|} = \alpha\vec{CX}$$
I have tried to inject the second formula in the first, but I wasn't able to figure out the necessary algebra, if that is the correct way.
Solution:
Using Shifrin's notation (see best answer below), we try to rearrange the sides to make easier to see the coefficients of $\vec{a}$ and $\vec{b}$, which must be equal on both sides of equation:
$$
\dfrac{\vec{a}}{\|\vec{a}\|} + \dfrac{\vec{b}}{\|\vec{b}\|} = \alpha\vec{a} + \alpha t (\vec{b} - \vec{a})
$$
$$
\dfrac{\vec{a}}{\|\vec{a}\|} + \dfrac{\vec{b}}{\|\vec{b}\|} = (\alpha - \alpha t)\vec{a} + \alpha t \vec{b}
$$
Hence, $\dfrac{1}{\|\vec{a}\|} = (\alpha - \alpha t) \iff \|\vec{a}\| = \dfrac{1}{(\alpha - \alpha t)}$ ,where $ \|\vec{a}\| \neq 0 $
$\dfrac{1}{\|\vec{b}\|} = \alpha t \iff \|\vec{b}\| = \dfrac{1}{\alpha t}$ ,where $ \|\vec{b}\| \neq 0 $
From these, $\cfrac{\|\vec{CA}\|}{\|\vec{AX}\|} = \cfrac{\cfrac{1}{(\alpha - \alpha t)}}{\|t(\vec{b} - \vec{a})\|} = \cfrac{1}{\|\alpha t(1 - t)(\vec{b} - \vec{a})\|}$
and $\cfrac{\|\vec{CB}\|}{\|\vec{BX}\|} = \cfrac{\cfrac{1}{\alpha t}}{\|(t - 1)(\vec{b} - \vec{a})\|} = \cfrac{1}{\|\alpha t(1 - t)(\vec{b} - \vec{a})\|}$
Of course you left out the crucial hypothesis that $\overrightarrow{CX}$ bisects $\angle ACB$. (But that's what your displayed equation is equivalent to.)
HINT: You wrote down one condition. There's also another: $X$ lies on $\overline{AB}$. Let $\vec a = \overrightarrow{CA}$ and $\vec b = \overrightarrow{CB}$ and write both equations in terms of $\vec a$ and $\vec b$.
Because $\vec a$ and $\vec b$ are not parallel, if you get $s\vec a + t\vec b = u\vec a + v \vec b$, what can you deduce about the scalars $s,t,u,v$?