Talk:Steiner-Lehmus Theorem

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--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)