telegrafyx

2022-06-22

In $ABC$, $\mathrm{\angle }B<\mathrm{\angle }C$. If $D$ and $E$ are in $AC$ and $AB$ respectively, such that $BD$ and $CE$ are angle bisectors, prove that $BD>CE$.
Well, $AC, but is there a formula for angle bisector length?

Eleanor Luna

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:
$\left\{\begin{array}{l}x+y=a\\ \frac{x}{y}=\frac{b}{c}\end{array}$
Solving this is easy and it yields:
$x=\frac{ab}{b+c}\phantom{\rule{1em}{0ex}},y=\frac{ac}{b+c}.$
Finally, Stewart's theorem gives us:
$a\cdot {l}_{a}^{2}=\frac{{b}^{2}ac+{c}^{2}ab}{b+c}-\frac{{a}^{3}bc}{\left(b+c{\right)}^{2}}$
or
${l}_{a}=\sqrt{bc-\frac{{a}^{2}bc}{\left(b+c{\right)}^{2}}}=\frac{2\sqrt{bcp\left(p-a\right)}}{b+c}$
where $p=\frac{a+b+c}{2}$ is the semiperimeter.

With this formula in hand and in the setup of your problem, one can immediately see that:
$C{E}^{2}-B{D}^{2}={l}_{c}^{2}-{l}_{b}^{2}=ab-\frac{{c}^{2}ab}{\left(a+b{\right)}^{2}}-ac+\frac{{b}^{2}ac}{\left(a+c{\right)}^{2}}=$
$=a\left(b-c\right)\left(1+\frac{bc\left({a}^{2}+{b}^{2}+{c}^{2}+bc+2ab+2ac\right)}{\left(a+b{\right)}^{2}\left(a+c{\right)}^{2}}\right)\le 0.$
Thus, ${l}_{b}\ge {l}_{c}.$

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