Angle bisector comparison

Question

In $ABC$, $\angle{B}<\angle{C}$. If $D$ and $E$ are in $AC$ and $AB$ respectively, such that $BD$ and $CE$ are angle bisectors, prove that $BD > CE$.

Well, $AC < AB$, but is there a formula for angle bisector length?

Answer 1

Geometry of a triangle is studied extensively. Let the sides be $a,b,c$ respectively and $l_a$ be the length of angle bisector from vertex $A$ onto the side whose length is $a.$ The Angle Bisector property tells us that $l_a$ bisects the side $BC$ in the proportion $AC:AB =b:c.$ This gives us two equations: $$\begin{cases} x+y = a\\ \dfrac{x}{y} = \dfrac{b}{c} \end{cases}$$ Solving this is easy and it yields: $$x = \dfrac{ab}{b+c}\quad , y = \dfrac{ac}{b+c}.$$ Finally, Stewart's theorem gives us: $$a\cdot l_a^2 = \dfrac{b^2ac+c^2ab}{b+c} - \dfrac{a^3bc}{(b+c)^2}$$ or $$l_a = \sqrt{bc - \dfrac{a^2bc}{(b+c)^2}} = \dfrac{2\sqrt{bcp(p-a)}}{b+c}$$ where $p = \dfrac{a+b+c}{2}$ is the semiperimeter.

With this formula in hand and in the setup of your problem, one can immediately see that: $$CE^2 - BD^2 = l_c^2 - l_b^2 = ab-\dfrac{c^2ab}{(a+b)^2} - ac+\dfrac{b^2ac}{(a+c)^2} = $$ $$=a(b-c)\left(1 + \dfrac{bc(a^2+b^2+c^2+bc+2ab+2ac)}{(a+b)^2(a+c)^2}\right)\leq 0.$$ Thus, $l_b\geq l_c.$