Let triangle ABC (denoted $\triangle ABC$) be an isosceles triangle in the plane where $|AB|=|AC|$. We try to prove that the perpendicular bisector of the segment $BC$ coincides with the bisector of $\angle A$ (i.e. they are the same line when extend) and that it cuts the line $BC$ in half. The diagram shows the configuration of $\triangle ABC $.
My proof is as follows.
By definition of the angle bisector of $\angle A$, we have $\angle BAO = \angle CAO =\frac{1}{2}\angle A.$ Let the bisector of $\angle A$ intersect the line segment $BC$ at the point $O.$ As $\triangle ABC $ is isosceles, we can say that $\angle OBA =\angle ACO = \alpha \text{ (say) }, $ (assume without proof) and so by side angle side (SAS) (assume without proof) we have shown that $\triangle OBA \cong \triangle OCA.$ This means that $|BO|=|OC|$ and that $\angle COB = \angle BOA=$ since both $\angle COB =\angle BOA$ and the fact that $\angle COB + \angle BOA=\pi, $ ( since $\angle COB $ is a straight angle) we conclude that $\angle AOB =\angle COB =\pi /2 .$ So we have shown that the bisector of $\angle A $ is actually the same line as the perpendicular bisector of the segment $AC$ if both lines are extended infinitely. $\ \ \ \ \square $
Is this proof valid, or are there any mistakes or parts i'm missing?
The proof is perfectly fine. However, there is a simpler proof with congruent triangles. Additionally, the similar side lengths of the big triangle could equal x. With this, control a function for an angle $\theta $, and you should be fine.