dumnealorjavgj

## Answered question

2022-04-06

How do I prove that a triangle with sides a, b, c, has an angle bisector (bisecting angle A) is of length:
$\frac{2\sqrt{bcs\left(s-a\right)}}{b+c}$
I have tried using the sine and cosine rule but have largely failed. A few times I have found a way but they are way too messy to work with.

### Answer & Explanation

reflam2kfnr

Beginner2022-04-07Added 16 answers

A method where no trigonometry is used.

Consider triangle $ABC$. Let $AD$, the angle bisector, intersect the circumcircle at $L$. Join $LC$. Consider triangle $ABD$ and triangle $ALC$.

Triangle $ABD$ is similar to triangle $ALC$ (by A.A similarity theorem). Therefore,
$\frac{AD}{AC}=\frac{AB}{AL}$
i.e,
$AD\cdot AL=AC\cdot AB$
$=AD\left(AD+DL\right)=AC\cdot AB$

By power of point of point result:
$AD\cdot DL=BD\cdot DC$
$BD=BC\cdot \frac{AB}{AB+AC}$
$DC=BC\cdot \frac{AC}{AB+AC}$
In (1) ,
$AD\cdot AD=AC\cdot AB-B{C}^{2}\cdot AB\cdot \frac{AC}{\left(AB+AC{\right)}^{2}}$
$A{D}^{2}=AC\cdot AB\left(1-\frac{B{C}^{2}}{\left(AB+AC{\right)}^{2}}\right)$
If $AB=c,BC=a,AC=b$:
$A{D}^{2}=bc\left(1-\frac{{a}^{2}}{\left(b+c{\right)}^{2}}\right)$

Berghofaei0e

Beginner2022-04-08Added 3 answers

Assuming the usage of Trigonometry is allowed,

Let $AD$ be the bisector of $\mathrm{\angle }BAC$

$\mathrm{△}ABC=\frac{1}{2}bc\mathrm{sin}A$
$\mathrm{△}ABD=\frac{1}{2}\cdot c\cdot AD\mathrm{sin}\frac{A}{2}$ and $\mathrm{△}ADC=\frac{1}{2}\cdot b\cdot AD\mathrm{sin}\frac{A}{2}$
$\mathrm{△}ABC=\mathrm{△}ABD+\mathrm{△}ADC$
$\mathrm{sin}A=2\mathrm{sin}\frac{A}{2}\mathrm{cos}\frac{A}{2}$
As $00\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}\frac{A}{2}=+\sqrt{\frac{1+\mathrm{cos}A}{2}}$ as $\mathrm{cos}A=2{\mathrm{cos}}^{2}\frac{A}{2}-1$
Use $\mathrm{cos}A=\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}$ and $2s=a+b+c$

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