The following is my proof of the statement Theorem: In $\triangle ABC$, $|\overline{AC}|=|\overline{BC}|$ $\iff$ $|\overline{AP’}|=|\overline{BP}|$ where $\overline{AP’}$ is the angle bisector of $\angle A$ and $\overline{BP}$ is the angle bisector of $\angle B$. I am just looking for verification that this is correct, or any proof by a more elegant method. I produced this proof without using any online resources so I apologize if this is a duplicate question.
Proof: First Direction, $|\overline{AC}|=|\overline{BC}|$, $\angle \theta = \angle \theta$, and $\angle \theta' = \angle \theta'$ by hyp. Then $2 \theta = 2\theta' \implies \theta = \theta'$ by isosceles, and $\triangle AP'B \cong \triangle BPA$ by ASA. Therefore $|\overline{AP’}|=|\overline{BP}|$ $\blacksquare$
Second Direction, $|\overline{AP’}|=|\overline{BP}|$, $\angle \theta = \angle \theta$, and $\angle \theta' = \angle \theta'$ by hyp. Construct $\overline{PP'}$. The remainder of the proof involves relating both properties of the givens through law of signs, equations are numbered for convenience:
$$1. \triangle PDP': \frac{\sin{\gamma}}{PP'}=\frac{\sin{\beta}}{DP'}=\frac{\sin{\beta'}}{DP}$$
$$ 2.\triangle APP': \frac{\sin{\theta}}{PP'}=\frac{\sin{(\alpha + \beta)}}{AP'}=\frac{\sin{\beta'}}{AP}, 3. \triangle BP'P: \frac{\sin{\theta'}}{PP'}=\frac{\sin{(\alpha' + \beta')}}{BP}=\frac{\sin{\beta}}{BP'}$$
$$4. \triangle AP'B: \frac{\sin{\theta}}{BP'}=\frac{\sin{2\theta'}}{AP'}=\frac{\sin{\alpha'}}{AB}, 5. \triangle APB: \frac{\sin{\theta'}}{AP}=\frac{\sin{2\theta}}{BP}=\frac{\sin{\alpha}}{AB}$$
$$\frac{1.}{2.}: \frac{\sin{\gamma}}{\sin{\theta}}=\frac{AP}{DP}, \frac{1.}{3.}: \frac{\sin{\gamma}}{\sin{\theta'}}=\frac{BP'}{DP} \implies \frac{\sin{\theta'}}{\sin{\theta}}=\frac{AP}{BP'}$$
$$\frac{4.}{5.}: \frac{\sin{\theta'}}{\sin{\theta}}=\frac{AP}{BP'}\frac{\sin{2\theta}}{\sin{2\theta'}} \implies \sin{2\theta} = \sin{2\theta}$$
This fact implies either that $2\theta = 2\theta'$ or $2\theta + 2\theta' = 180°$ but if the latter was true then $\mu = 0$ and that is absurd. Therefore $2\theta = 2\theta'$ and by the Isosceles Angle Theorem $|\overline{AC}| = |\overline{BC}| \blacksquare$