Steiner-Lehmus Theorem/Proof 2
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:
| \(\ds \omega_\alpha^2\) | \(=\) | \(\ds b c \paren {1 - \dfrac {a^2} {\paren {b + c}^2} }\) | ||||||||||||
| \(\ds \omega_\beta^2\) | \(=\) | \(\ds a c \paren {1 - \dfrac {b^2} {\paren {a + c}^2} }\) |
Equating $\omega_\alpha^2$ with $\omega_\beta^2$ yields:
| \(\ds b c - \dfrac {b c a^2} {\paren {b + c}^2}\) | \(=\) | \(\ds a c - \dfrac {a c b^2} {\paren {a + c}^2}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds c \paren {b - a}\) | \(=\) | \(\ds \dfrac {b c a^2} {\paren {b + c}^2} - \dfrac {a c b^2} {\paren {a + c}^2}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds c \paren {b - a} \paren {b + c}^2 \paren {a + c}^2\) | \(=\) | \(\ds b c a^2 \paren {a + c}^2 - a c b^2 \paren {b + c}^2\) | |||||||||||
| \(\text {(1)}: \quad\) | \(\ds \) | \(=\) | \(\ds c a \paren {a \paren {a + c}^2 - b\paren {b + c}^2}\) |
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 $(1)$.
Aiming for a contradiction, suppose $b \ne a$.
Without loss of generality, suppose $b > a$.
Because $c \paren {b - a} \paren {b + c}^2 \paren {a + c}^2 > 0$, the left hand side of $(1)$ is positive.
We have by hypothesis that:
- $a$, $b$, and $c$ are positive real numbers
and by hypothesis:
- $a < b$
By Real Number Ordering is Compatible with Multiplication and since $a > 0$:
- $a^2 < a \cdot b$
For the same reason and since $b > 0$:
- $a \cdot b < b^2$
By Real Number Ordering is Transitive:
- $a^2 < b^2$
Then:
| \(\ds a^3\) | \(<\) | \(\ds b^3\) | mutatis mutandis | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a^2 - b^2\) | \(<\) | \(\ds 0\) | |||||||||||
| \(\ds a^3 - b^3\) | \(<\) | \(\ds 0\) |
Since $c a > 0$, the right hand side of $(1)$ will be negative if and only if:
- $a \paren {a + c}^2 - b \paren {b + c}^2 < 0$
We rearrange as follows:
| \(\ds a \paren {a + c}^2 - b \paren {b + c}^2\) | \(=\) | \(\ds a \paren {a^2 + 2 a c + c^2} - b \paren {b^2 + 2 b c + c^2}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds a^3 + 2 a^2 c + a c^2 - b^3 - 2 b^2 c - b c^2\) | ||||||||||||
| \(\text {(2)}: \quad\) | \(\ds \) | \(=\) | \(\ds \paren {a^3 - b^3} + 2 c \paren {a^2 - b^2} + c^2 \paren {a - b}\) |
From above, each term in parenthesis in $(2)$ is negative.
As $c > 0$ we have:
| \(\ds 2 c\) | \(>\) | \(\ds 0\) | ||||||||||||
| \(\ds c^2\) | \(>\) | \(\ds 0\) |
Hence each term in $(2)$ is negative.
Hence their sum is also negative.
Therefore, the right hand side of $(1)$ is negative.
Hence we have reached a contradiction.
$\Box$
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 $(1)$.
Therefore, by Proof by Contradiction, $b = a$ is the only solution for $(1)$.
Therefore $ABC$ is an isosceles triangle.
$\blacksquare$
Source of Name
This entry was named for Jakob Steiner and Daniel Christian Ludolph Lehmus.