User talk:Jhoshen1/Sandbox

Proof 2
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:
 * $\omega _\alpha ^2 = bc \paren { 1 - \dfrac {  a^2 } {\paren { b + c}^2 } }$


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

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


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


 * $\leadsto c(b - a) = \dfrac { {b c a^2 }} {\paren { b + c}^2 } - \dfrac { {a c b^2 }} {\paren {a + c}^2 } $
 * $\leadsto c(b - a)(b + c)^2 (a + c)^2 = bca^2 (a + c)^2  - acb^2 (b + c)^2 $
 * $\leadsto c(b - a)(b + c)^2 (a + c)^2 = ca \paren { a(a + c)^2  - b(b + c)^2 } \;\;\; (1) $

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 $(1)$. We do so by assuming that $b>a$. Because $ c(b - a)(b + c)^2 (a + c)^2 > 0$, the left hand side of $(1)$ is positive. By assumption $b > a $ and $b+c > a+c $
 * $\leadsto ca \paren { a(a + c)^2 < b(b + c)^2 } $,

Therefore, the right hand side of $(1)$ is negative. And we have a contradiction. Similarly, we also get a contradiction by assuming $b< a$. This assumption yields a negative left hand side and a positive right hand side for $(1)$. To sum up, $b=a$ is the only solution for $(1)$

Therefore $ABC$ is an isosceles triangle.