2
$\begingroup$

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 $.

enter image description here

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?

$\endgroup$
  • $\begingroup$ It seems that you made an error - the bisector is of A $\endgroup$ – Moti Oct 22 '16 at 1:20
  • $\begingroup$ Apart of this the proof seems OK $\endgroup$ – Moti Oct 22 '16 at 1:22
  • $\begingroup$ 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. $\endgroup$ – Anon Oct 22 '16 at 12:15
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.