Steiner-Lehmus Theorem/Proof 1
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Theorem
Let $ABC$ be a triangle.
Denote the lengths of the angle bisectors through the vertices $A$ and $B$ by $\omega_\alpha$ and $\omega_\beta$.
Let $\omega_\alpha = \omega_\beta$.
Then $ABC$ is an isosceles triangle.
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:
- $\omega_\alpha^2 = \dfrac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}$
- $\omega_\beta^2 = \dfrac {a c} {\paren {a + c}^2} \paren {\paren {a + c}^2 - b^2}$
Equating $\omega_\alpha^2$ with $\omega_\beta^2$:
| \(\ds \dfrac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}\) | \(=\) | \(\ds \dfrac {a c} {\paren {a + c}^2} \paren {\paren {a + c}^2 - b^2}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds b c \paren {a + c}^2 \paren {\paren {b + c}^2 - a^2}\) | \(=\) | \(\ds a c \paren {b + c}^2 \paren {\paren {a + c}^2 - b^2}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds b c \paren {a + c}^2 \paren {b + c + a} \paren {b + c - a}\) | \(=\) | \(\ds a c \paren {b + c}^2 \paren {a + c + b} \paren {a + c - b}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds b \paren {a + c}^2 \paren {b + c - a}\) | \(=\) | \(\ds a \paren {b + c}^2 \paren {a + c - b}\) | as $a + b + c > 0$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a^2 b^2 + 2 a b^2 c + b^2 c^2 + a^2 b c + 2 a b c^2 + b c^3 - a^3 b - 2 a^2 b c - a b c^2\) | \(=\) | \(\ds a^2 b^2 + 2 a^2 b c + a^2 c^2 + a b^2 c + 2 a b c^2 + a c^3 - a b^3 - 2 a b^2 c - a b c^2\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 2 a b^2 c + b^2 c^2 - a^2 b c + b c^3 - a^3 b\) | \(=\) | \(\ds 2 a^2 b c + a^2 c^2 - a b^2 c + a c^3 - a b^3\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 3 a b^2 c - 3 a^2 b c + b^2 c^2 + b c^3 - a^3 b - a^2 c^2 - a c^3 + a b^3\) | \(=\) | \(\ds 0\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 3 a b c \paren {b - a} + c^2 \paren {b^2 - a^2} + c^3 \paren {b - a} + a b \paren {b^2 - a^2}\) | \(=\) | \(\ds 0\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {b - a} \paren {3 a b c + a b \paren {a + b} + c^3 + c^2 \paren {a + b} }\) | \(=\) | \(\ds 0\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds b - a\) | \(=\) | \(\ds 0\) | as $a, b, c > 0$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a\) | \(=\) | \(\ds b\) |
Therefore $ABC$ is an isosceles triangle.
$\blacksquare$
Source of Name
This entry was named for Jakob Steiner and Daniel Christian Ludolph Lehmus.