Steiner-Lehmus Theorem/Proof 2

Proof
Let $a$, $b$, and $c$ be the sides opposite $A$, $B$ and $C$ respectively.

By Length of Angle Bisector, $\omega_\alpha, \omega_\beta$ are given by:
 * $\omega_\alpha^2 = b c \paren {1 - \dfrac {a^2} {\paren {b + c}^2} }$


 * $\omega_\beta^2 = a c \paren {1 - \dfrac {b^2} {\paren {a + c}^2} }$

Equating $\omega_\alpha^2$ with $\omega_\beta^2$ yields:


 * $b c - \dfrac {b c a^2} {\paren {b + c}^2} = a c - \dfrac {a c b^2} {\paren {a + c}^2}$

Substituting $b$ for $a$ in $(1)$ proves that $b = a$ is a solution for $(1)$.

We still have to show that $b = a$ is the only solution for $(1)$.

$b \ne a$.

, suppose $b > a$.

Because $c \paren {b - a} \paren {b + c}^2 \paren {a + c}^2 > 0$, the of $(1)$ is positive.

Hence:


 * $\leadsto c a \paren {a \paren {a + c}^2 < b \paren {b + c}^2}$

Therefore, the of $(1)$ is negative.

Hence we have reached a contradiction.

Similarly, we also get a contradiction by assuming $b < a$.

This assumption yields a negative and a positive  for $(1)$.

Hence, by Proof by Contradiction, $b = a$ is the only solution for $(1)$.

Therefore $ABC$ is an isosceles triangle.