Steiner-Lehmus Theorem/Lemma 2

Lemma for Steiner-Lehmus Theorem

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:

$CN < BM$

Proof

Let $\angle NBM = \angle MBC = \beta$.

Let $\angle NCB = \angle NCM = \gamma$.

We have by hypothesis that:

$\angle ABC < \angle ACB$

Hence:

$\beta < \gamma$

Let a circle be drawn through $BCN$.

Let $M'$ be on the circle such that:

$\angle NCM' = \beta$

Since $\beta < \gamma$:

$\angle NCM' < \angle NCM$

and so:

$BM' < BM$

By Lemma $1$, since $\angle NBC < \angle M'CB$:

$CN < BM'$

Hence:

$CN < BM' < BM$

Thus:

$CN < BM$

$\blacksquare$