I proved this property with an approach involving vectors. However, there should be a much simpler, elegant geometric proof, probably utilising a bunch of angles. Here is a diagram exemplifying the property:
And here is my very tedious approach:
Let $ \lambda_1 = \frac{ \overrightarrow{OA} \cdot \overrightarrow{OB}}{\overrightarrow{OA} \cdot \overrightarrow{OA}} = \frac{|\overrightarrow{OB}|}{|\overrightarrow{OA}|} \cos(\theta) $ and $ \lambda_2 = \frac{ \overrightarrow{OA} \cdot \overrightarrow{OB}}{\overrightarrow{OB} \cdot \overrightarrow{OB}} = \frac{|\overrightarrow{OA}|}{|\overrightarrow{OB}|} \cos(\theta) $
By considering the intersection of the lines $BP_1$ and $AP_2$, where
$$ AP_2 := (1-\mu_1)\lambda_1 \overrightarrow{OA} + \mu_1 \overrightarrow{OB} \\ BP_1 := (1-\mu_2)\lambda_2 \overrightarrow{OB} + \mu_2 \overrightarrow{OA} \\ (1-\mu_1)\lambda_1 \overrightarrow{OA} + \mu_1 \overrightarrow{OB} = (1-\mu_2)\lambda_2 \overrightarrow{OB} + \mu_2 \overrightarrow{OA} \\ \implies ((1-\mu_1)\lambda_1 - \mu_2 )\overrightarrow{OA} = ((1-\mu_2)\lambda_2 - \mu_1 )\overrightarrow{OB} \\ \implies (1-\mu_1)\lambda_1 = \mu_2 \quad \text{ and } \quad (1-\mu_2)\lambda_2 = \mu_1 \\ \implies \mu_1 = \frac{\lambda_2 - \lambda_1\lambda_2}{1-\lambda_1\lambda_2} \quad \text{ and } \quad \mu_2 = \frac{\lambda_1 - \lambda_1\lambda_2}{1-\lambda_1\lambda_2} \\ \ \\ \ \\ \implies \overrightarrow{OX} = \frac{1}{1-\lambda_1\lambda_2} \left( (\lambda_1 - \lambda_1\lambda_2)\overrightarrow{OA} + (\lambda_2 -\lambda_1\lambda_2) \overrightarrow{OB} \right) $$
Similarly, by considering the intersection of $AB$ and $OX$,
$$ \overrightarrow{OP_3} = \frac{1}{\lambda_1 + \lambda_2 - 2\lambda_1\lambda_2} \left( (\lambda_1 - \lambda_1\lambda_2)\overrightarrow{OA} + (\lambda_2 -\lambda_1\lambda_2)\overrightarrow{OB} \right) \\ \overrightarrow{P_1P_3} = \frac{1-\lambda_1}{\lambda_1 + \lambda_2 - 2\lambda_1\lambda_2} \left( (\lambda_1 - 2\lambda_1\lambda_2)\overrightarrow{OA} + \lambda_2\overrightarrow{OB} \right) \\ \overrightarrow{P_2P_3} = \frac{1-\lambda_2}{\lambda_1 + \lambda_2 - 2\lambda_1\lambda_2} \left( \lambda_1\overrightarrow{OA} + (\lambda_2 - 2\lambda_1\lambda_2)\overrightarrow{OB} \right) \\ $$
Since we are looking for angles, the magnitudes of the vectors do not matter; we may strip the constants away to simplify our dot product.
$$ \overrightarrow{OP_{3\text{ dir}}} \cdot \overrightarrow{P_1P_{3\text{ dir}}} = |\overrightarrow{OP_{3\text{ dir}}}| |\overrightarrow{P_1P_{3\text{ dir}}}| \cos(\alpha_1) = \ldots = \sin^2(\theta)\cos^2(\theta) (\overrightarrow{OA} - \overrightarrow{OB})^2 \\ \overrightarrow{OP_{3\text{ dir}}} \cdot \overrightarrow{P_2P_{3\text{ dir}}} = |\overrightarrow{OP_{3\text{ dir}}}| |\overrightarrow{P_2P_{3\text{ dir}}}| \cos(\alpha_2) = \ldots = \sin^2(\theta)\cos^2(\theta) (\overrightarrow{OA} - \overrightarrow{OB})^2 \\ \implies \frac{\overrightarrow{OP_{3\text{ dir}}} \cdot \overrightarrow{P_1P_{3\text{ dir}}}}{\overrightarrow{OP_{3\text{ dir}}} \cdot \overrightarrow{P_2P_{3\text{ dir}}}} = \frac{|\overrightarrow{P_1P_{3\text{ dir}}}|\cos(\alpha_1)}{|\overrightarrow{P_2P_{3\text{ dir}}}|\cos(\alpha_2)} = 1 $$
Finally, by expansion, note that $\frac{|\overrightarrow{P_1P_{3\text{ dir}}}|}{|\overrightarrow{P_2P_{3\text{ dir}}}|} = \frac{\left| |\overrightarrow{OB}| \widehat{OA} - 2\overrightarrow{OA}\cos(\theta) + |\overrightarrow{OA}| \widehat{OB}\right|}{ \left| |\overrightarrow{OB}| \widehat{OA} - 2\overrightarrow{OB}\cos(\theta) + |\overrightarrow{OA} \widehat{OB}\right|} $, which can then be geometrically proved to be $\frac{|\overrightarrow{OA}-\overrightarrow{OB}|}{|\overrightarrow{OA}-\overrightarrow{OB}|} = 1$, here is another diagram:
Therefore, $\cos(\alpha_1) = \cos(\alpha_2) \implies \alpha_1=\alpha_2 $.
Now you see why I think there must be a much easier method. How else can we prove this?
