My question is clearly stated in the title.
By that theorem, I am not referring to the one saying “points on the angle bisector of an angle are equidistant from the sides”. It is the one that divides the opposite side instead.
The proof is not that difficult (I think!) but I find it was hardly mentioned. Even using “converse of angle bisector theorem” as search keyword in WIKI, the reply is “The page …. does not exist.”
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$\begingroup$You can find the proof in Classical Geometry: Euclidean, Transformational, Inversive, and Projective, page 144. It uses corollary 5.2.4:
If $B$, $C$, and $D$ are collinear points and if $A$ is a point not collinear with them, then $$\frac{[ABD]}{[ADC]}=\frac{BD}{DC}\;.$$
as follows:
So to be explicit about the question you ask, here is this converse theorem as I understand it:
Given a triangle $\triangle ABC$ and a point $D$ on $BC$ which satisfies $$\frac{\lvert BD\rvert}{\lvert DC\rvert}=\frac{\lvert AB\rvert}{\lvert AC\rvert}\;.$$ Then the line $AD$ will be a bisector of the angle $\angle CAB$.
Is this the theorem which you want to have verified?
Well, take the forward direction: starting at the angle bisector, you know there is a point $D'$ which satisfies the equation above. So the question you have to ask yourself is this: given one point which does satisfy the equation, can there be another?
You can consider the line $BC$ as a real projective axis, with $C$ as its zero point and $B$ as it's one point. So the condition
$$\frac{\lvert BD\rvert}{\lvert DC\rvert}=a$$
(where $a$ is fixed) translates into
$$\frac{\lvert 1-x\rvert}{\lvert x\rvert}=a$$
If you omit the absolute value signs, then either numerator and denominator have the same sign, in which case the right hand side should still read $a$. Or they have different signs, in which case the right hand side would become $-a$. So here you have
$$\frac{1-x}{x}=\pm a \qquad\Rightarrow\qquad x=\frac{1}{1\pm a}$$
So there is indeed a second solution, one where $x$ is between $0$ and $1$ and one where it is smaller than zero or greater than one. So there is a second point $D$ which is not equal to the $D'$ on the inner angle bisector, which does not lie on the segment $BC$, but which does lie on the extended line $BC$ and which does satisfy the above equation.
With a bit more work, again along the lines of the Wikipedia proof, you can show that this second point is the point where the line $BC$ intersects the exterior angle bisector.
So the theorem as stated above is true as long as you either restrict “point $D$ on $BC$” to mean “point $D$ on the line segment $BC$” or interpret the conclusion as “$AD$ will be an inner or outer bisector of the angle $\angle CAB$”.
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