dumnealorjavgj

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:

$\frac{2\sqrt{bcs(s-a)}}{b+c}$

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.

$\frac{2\sqrt{bcs(s-a)}}{b+c}$

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.

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,

$AD\cdot AL=AC\cdot AB$

$=AD(AD+DL)=AC\cdot AB$

$=AD\cdot AD+AD\cdot DL=AC\cdot AB\text{... (1)}$

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) ,

$AD\cdot AD=AC\cdot AB-B{C}^{2}\cdot AB\cdot \frac{AC}{(AB+AC{)}^{2}}$

$A{D}^{2}=AC\cdot AB{\textstyle (}1-\frac{B{C}^{2}}{(AB+AC{)}^{2}}{\textstyle )}$

If $AB=c,BC=a,AC=b$:

$A{D}^{2}=bc{\textstyle (}1-\frac{{a}^{2}}{(b+c{)}^{2}}{\textstyle )}$

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,

$AD\cdot AL=AC\cdot AB$

$=AD(AD+DL)=AC\cdot AB$

$=AD\cdot AD+AD\cdot DL=AC\cdot AB\text{... (1)}$

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) ,

$AD\cdot AD=AC\cdot AB-B{C}^{2}\cdot AB\cdot \frac{AC}{(AB+AC{)}^{2}}$

$A{D}^{2}=AC\cdot AB{\textstyle (}1-\frac{B{C}^{2}}{(AB+AC{)}^{2}}{\textstyle )}$

If $AB=c,BC=a,AC=b$:

$A{D}^{2}=bc{\textstyle (}1-\frac{{a}^{2}}{(b+c{)}^{2}}{\textstyle )}$

Berghofaei0e

Beginner2022-04-08Added 3 answers

Assuming the usage of Trigonometry is allowed,

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

$\mathrm{\u25b3}ABC=\frac{1}{2}bc\mathrm{sin}A$

$\mathrm{\u25b3}ABD=\frac{1}{2}\cdot c\cdot AD\mathrm{sin}\frac{A}{2}$ and $\mathrm{\u25b3}ADC=\frac{1}{2}\cdot b\cdot AD\mathrm{sin}\frac{A}{2}$

$\mathrm{\u25b3}ABC=\mathrm{\u25b3}ABD+\mathrm{\u25b3}ADC$

$\mathrm{sin}A=2\mathrm{sin}\frac{A}{2}\mathrm{cos}\frac{A}{2}$

As $0<A<\pi ,0<\frac{A}{2}<\frac{\pi}{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}\frac{A}{2}>0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}\frac{A}{2}=+\sqrt{\frac{1+\mathrm{cos}A}{2}}$ 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 $2s=a+b+c$

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

$\mathrm{\u25b3}ABC=\frac{1}{2}bc\mathrm{sin}A$

$\mathrm{\u25b3}ABD=\frac{1}{2}\cdot c\cdot AD\mathrm{sin}\frac{A}{2}$ and $\mathrm{\u25b3}ADC=\frac{1}{2}\cdot b\cdot AD\mathrm{sin}\frac{A}{2}$

$\mathrm{\u25b3}ABC=\mathrm{\u25b3}ABD+\mathrm{\u25b3}ADC$

$\mathrm{sin}A=2\mathrm{sin}\frac{A}{2}\mathrm{cos}\frac{A}{2}$

As $0<A<\pi ,0<\frac{A}{2}<\frac{\pi}{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}\frac{A}{2}>0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}\frac{A}{2}=+\sqrt{\frac{1+\mathrm{cos}A}{2}}$ 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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