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

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

Eleanor Luna

Beginner2022-06-23Added 19 answers

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:

$\{\begin{array}{l}x+y=a\\ {\displaystyle \frac{x}{y}}={\displaystyle \frac{b}{c}}\end{array}$

Solving this is easy and it yields:

$x={\displaystyle \frac{ab}{b+c}}\phantom{\rule{1em}{0ex}},y={\displaystyle \frac{ac}{b+c}}.$

Finally, Stewart's theorem gives us:

$a\cdot {l}_{a}^{2}={\displaystyle \frac{{b}^{2}ac+{c}^{2}ab}{b+c}}-{\displaystyle \frac{{a}^{3}bc}{(b+c{)}^{2}}}$

or

${l}_{a}=\sqrt{bc-{\displaystyle \frac{{a}^{2}bc}{(b+c{)}^{2}}}}={\displaystyle \frac{2\sqrt{bcp(p-a)}}{b+c}}$

where $p={\displaystyle \frac{a+b+c}{2}}$ is the semiperimeter.

With this formula in hand and in the setup of your problem, one can immediately see that:

$C{E}^{2}-B{D}^{2}={l}_{c}^{2}-{l}_{b}^{2}=ab-{\displaystyle \frac{{c}^{2}ab}{(a+b{)}^{2}}}-ac+{\displaystyle \frac{{b}^{2}ac}{(a+c{)}^{2}}}=$

$=a(b-c)(1+{\displaystyle \frac{bc({a}^{2}+{b}^{2}+{c}^{2}+bc+2ab+2ac)}{(a+b{)}^{2}(a+c{)}^{2}}})\le 0.$

Thus, ${l}_{b}\ge {l}_{c}.$

$\{\begin{array}{l}x+y=a\\ {\displaystyle \frac{x}{y}}={\displaystyle \frac{b}{c}}\end{array}$

Solving this is easy and it yields:

$x={\displaystyle \frac{ab}{b+c}}\phantom{\rule{1em}{0ex}},y={\displaystyle \frac{ac}{b+c}}.$

Finally, Stewart's theorem gives us:

$a\cdot {l}_{a}^{2}={\displaystyle \frac{{b}^{2}ac+{c}^{2}ab}{b+c}}-{\displaystyle \frac{{a}^{3}bc}{(b+c{)}^{2}}}$

or

${l}_{a}=\sqrt{bc-{\displaystyle \frac{{a}^{2}bc}{(b+c{)}^{2}}}}={\displaystyle \frac{2\sqrt{bcp(p-a)}}{b+c}}$

where $p={\displaystyle \frac{a+b+c}{2}}$ is the semiperimeter.

With this formula in hand and in the setup of your problem, one can immediately see that:

$C{E}^{2}-B{D}^{2}={l}_{c}^{2}-{l}_{b}^{2}=ab-{\displaystyle \frac{{c}^{2}ab}{(a+b{)}^{2}}}-ac+{\displaystyle \frac{{b}^{2}ac}{(a+c{)}^{2}}}=$

$=a(b-c)(1+{\displaystyle \frac{bc({a}^{2}+{b}^{2}+{c}^{2}+bc+2ab+2ac)}{(a+b{)}^{2}(a+c{)}^{2}}})\le 0.$

Thus, ${l}_{b}\ge {l}_{c}.$