Let triangle ABC (denoted △ABC\triangle ABC) be an isosceles triangle in the plane where |AB|=|AC||AB|=|AC|. We try to prove that the perpendicular bisector of the segment BCBC coincides with the bisector of ∠A\angle A (i.e. they are the same line when extend) and that it cuts the line BCBC in half. The diagram shows the configuration of △ABC\triangle ABC .

My proof is as follows.

By definition of the angle bisector of ∠A\angle A, we have ∠BAO=∠CAO=12∠A.\angle BAO = \angle CAO =\frac{1}{2}\angle A. Let the bisector of ∠A\angle A intersect the line segment BCBC at the point O.O.

As △ABC\triangle ABC is isosceles, we can say that ∠OBA=∠ACO=α (say) ,\angle OBA =\angle ACO = \alpha \text{ (say) }, (assume without proof) and so by side angle side (SAS) (assume without proof) we have shown that △OBA≅△OCA.\triangle OBA \cong \triangle OCA.

This means that |BO|=|OC||BO|=|OC| and that ∠COB=∠BOA=\angle COB = \angle BOA= since both ∠COB=∠BOA\angle COB =\angle BOA and the fact that ∠COB+∠BOA=π,\angle COB + \angle BOA=\pi, ( since ∠COB\angle COB is a straight angle) we conclude that ∠AOB=∠COB=π/2.\angle AOB =\angle COB =\pi /2 .

So we have shown that the bisector of ∠A\angle A is actually the same line as the perpendicular bisector of the segment ACAC if both lines are extended infinitely. ◻\ \ \ \ \square

Is this proof valid, or are there any mistakes or parts i’m missing?

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It seems that you made an error – the bisector is of A

– Moti

2 days ago

Apart of this the proof seems OK

– Moti

2 days ago

Yeah it’s because when I wrote the proof myself the vertices were ordered differently but I just pulled this picture of an isosceles triangle off the internet.

– Ben

yesterday

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