Steiner-Lehmus Theorem

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 1

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.

$\blacksquare$

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:

\(\ds \omega_\alpha^2\)

\(=\)

\(\ds b c \paren {1 - \dfrac {a^2} {\paren {b + c}^2} }\)

\(\ds \omega_\beta^2\)

\(=\)

\(\ds a c \paren {1 - \dfrac {b^2} {\paren {a + c}^2} }\)

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.

$\blacksquare$

Proof 3

Draw $DF \parallel BE$ and $EF \parallel BD$.

By Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel:

$\Box BEFD$ is a parallelogram.

$\leadsto \angle DFE = \beta$

Draw $FA$.

Let $\angle EFA = \gamma$.

Let $\angle EAF = \delta$.

Aiming for a contradiction, suppose $\alpha > \beta$.

Compare $\triangle BAD$ with $\triangle ABE$.

The base is the same, $BA$.

\(\ds BE\)

\(=\)

\(\ds AD\)

by hypothesis

\(\ds \alpha\)

\(>\)

\(\ds \beta\)

by hypothesis

\(\ds \leadsto \ \ \)

\(\ds BD\)

\(>\)

\(\ds AE\)

Greater Angle of Triangle Subtended by Greater Side

\(\ds BD\)

\(=\)

\(\ds EF\)

Definition of Parallelogram

\(\ds \leadsto \ \ \)

\(\ds EF\)

\(>\)

\(\ds AE\)

\(\ds \leadsto \ \ \)

\(\ds \delta\)

\(>\)

\(\ds \gamma\)

Greater Side of Triangle Subtends Greater Angle

\(\ds \alpha + \delta\)

\(>\)

\(\ds \beta + \gamma\)

Addition of Inequalities

\(\ds \leadsto \ \ \)

\(\ds AD\)

\(>\)

\(\ds DF\)

Greater Angle of Triangle Subtended by Greater Side

\(\ds DF\)

\(=\)

\(\ds BE\)

Definition of Parallelogram

\(\ds \leadsto \ \ \)

\(\ds AD\)

\(>\)

\(\ds BE\)

But this is a contradiction.

Hence by Proof by Contradiction it cannot be the case that $\alpha > \beta$.

$\Box$

Aiming for a contradiction, suppose $\beta > \alpha$.

Again compare $\triangle BAD$ with $\triangle ABE$.

They have the same base, $BC$.

\(\ds BE\)

\(=\)

\(\ds AD\)

by hypothesis

\(\ds \beta\)

\(>\)

\(\ds \alpha\)

by hypothesis

\(\ds \leadsto \ \ \)

\(\ds AE\)

\(>\)

\(\ds BD\)

Greater Angle of Triangle Subtended by Greater Side

\(\ds BD\)

\(=\)

\(\ds EF\)

Definition of Parallelogram

\(\ds \leadsto \ \ \)

\(\ds AE\)

\(>\)

\(\ds EF\)

\(\ds \leadsto \ \ \)

\(\ds \gamma\)

\(>\)

\(\ds \delta\)

Greater Side of Triangle Subtends Greater Angle

\(\ds \beta + \gamma\)

\(>\)

\(\ds \alpha + \delta\)

Addition of Inequalities

\(\ds \leadsto \ \ \)

\(\ds AD\)

\(>\)

\(\ds DF\)

Greater Angle of Triangle Subtended by Greater Side

\(\ds DF\)

\(=\)

\(\ds BE\)

Definition of Parallelogram

\(\ds \leadsto \ \ \)

\(\ds AD\)

\(>\)

\(\ds BE\)

But this is a contradiction.

Hence by Proof by Contradiction it cannot be the case that $\alpha > \beta$.

$\Box$

Since neither $\alpha > \beta$ nor $\alpha < \beta$, it must be the case that

$\alpha = \beta$

So by addition:

$\angle ABC = \angle BAC$

and by Triangle with Two Equal Angles is Isosceles:

$\triangle ABC$ is isosceles.

The result follows.

$\blacksquare$

Proof 4

Lemma $1$

In the same circle, let one chord $PR$, be larger than another, $PQ$.

Then the inscribed angle subtended by $PR$, with its vertex in the major arc, is larger.

$\Box$

Lemma $2$

Let $\triangle ABC$ be a triangle.

Let $\angle ABC$ be bisected by $BM$.

Let $\angle ACB$ be bisected by $CN$.

Let $\angle ABC < \angle ACB$.

Then:

$CN < BM$

$\Box$

Let $\triangle ABC$ be a triangle.

Let $\angle ABC$ be bisected by $BM$

Let $\angle ACB$ be bisected by $CN$.

Let $BM = CN$.

Suppose $\angle ABC < \angle ACB$.

By Lemma $2$:

$CN < BM$

This is a contradiction.

$\Box$

Suppose $\angle ABC < \angle ACB$.

By Lemma $2$:

$CN > BM$

This is a contradiction.

$\Box$

Since $\angle ABC$ can be neither less than nor greater than $\angle ACB$, $\angle ABC = \angle ACB$.

By Triangle with Two Equal Angles is Isosceles:

$\triangle ABC$ is an isosceles triangle.

The result follows.

$\blacksquare$

Converse

The converse of the theorem is also true, and is much easier to prove:

Let $ABC$ be an isosceles triangle with $BA = BC$.

Let $D$ be the point of intersection of the angle bisectors through the vertices $A$ with the side $BC$.

Likewise, let $E$ be the point of intersection of the angle bisectors through the vertices $C$ with the side $BA$.

Then:

$AD = CE$

Source of Name

This entry was named for Jakob Steiner and Daniel Christian Ludolph Lehmus.