Talk:Steiner-Lehmus Theorem

--Jhoshen1 (talk) 20:46, 29 January 2021 (UTC)The formula for the length of the vertex B angular bisector given in the proof is incorrect as it leads to the conclusion that $b=c$ instead of $b=a$. The formula given in the current proof for $\omega_\beta^2$ is:
 * $\omega_\beta^2 = \dfrac {b a} {\paren {b + a}^2} \paren {\paren {b + a}^2 - c^2}$

It should have been
 * $\omega_\beta^2 = \dfrac {c a} {\paren {c + a}^2} \paren {\paren {c + a}^2 - b^2}$

Shortly, I will be proposing a correction, a shortening and a simplification of the proof


 * While you may feel free to correct any mistakes in the existing proof, please do not replace it with your own proof (even if it is shorter and perhaps better).


 * Please add your new proof as a completely new section.


 * Many thanks for your co-operation. --prime mover (talk) 22:01, 29 January 2021 (UTC)


 * Now I've looked at it, it merely means that the second formula is for $\omega_\gamma^2$, where $\omega_\gamma$ is the angle bisector of $C$ rather than of $B$.


 * Hence it shows that $a = c$. --prime mover (talk) 22:04, 29 January 2021 (UTC)

--Jhoshen1 (talk) 00:51, 30 January 2021 (UTC):I need help to create a second proof to the theorem (I am new to this form).


 * The best way to learn is to do. --prime mover (talk) 10:00, 30 January 2021 (UTC)


 * There are 2 options for correcting the proof; either by restating the the theorem to

Let $ABC$ be a triangle.

Denote the lengths of the angle bisectors through the vertices $A$ and $C$ by $\omega_\alpha$ and $\omega_\gamma$.

Let $\omega_\alpha = \omega_\gamma$.

Then $ABC$ is an isosceles triangle.

Or correct most of the equations to show that $a=b$

Which option is preferable?


 * I have worked through it myself, to save you the agony. --prime mover (talk) 10:00, 30 January 2021 (UTC)