geometrytriangles
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.
Considering areas of triangles gives you $$\text{Area($\triangle{ABC}$)}=\text{Area($\triangle{ABD}$)}+\text{Area($\triangle{ACD}$)},$$ i.e. $$\frac 12\times 4\times 3\times\sin60^\circ=\frac 12\times4\times AD\times\sin30^\circ+\frac 12\times 3\times AD\times \sin30^\circ.$$
An elemenraty solution: In the following figure $|BC|=a,|CA|=b,|AB|=c$ and $[CD]$ angle bisector. Let's draw square $CEDF$ and $|CE|=x$. So, $|BE|=a-x$ and $|CD|=x\sqrt2$. Now $\triangle ABC \sim \triangle DBE$ and $$\dfrac{b}{a}=\dfrac{x}{a-x} $$
Therefore $x=\dfrac{ab}{a+b}$ and $|CD|=\dfrac{ab\sqrt2}{a+b}.$
Best Answer
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 $(1)$ , $$AD\cdot AD = AC\cdot AB-BC^2\cdot AB\cdot \frac{AC}{(AB+AC)^2}$$ $$AD^2 = AC\cdot AB\Bigl(1-\frac{BC^2}{(AB+AC)^2}\Bigr)$$ If $AB = c , BC = a , AC = b$: $$AD^2 = bc\Bigl( 1-\frac{a^2}{(b+c)^2}\Bigr)$$ Hence proved.