Question from Triangles (Geometry) including angle bisector

Question

Assume here $AB>AC$:

1. If the bisector of $\angle A $ in triangle $ABC$ meets $BC$ at $U$, prove that $$AU^2 = \frac{bc(1-a^2)}{(b+c) ^2} $$
2. If the external bisector of $\angle A $ meet BC at U' then prove that

$$AU'^2 =bc\left(\frac{a^2}{(c-b) ^2-1}\right) $$

Answer 1

Let $BC=a, AC=b, AB=c, AU=l, \angle CAU=\angle BAU=\alpha, CD||AU'$.

$1.$ The area of $\Delta ABC$ is equal to the sum of $\Delta ABU$ and $\Delta ACU$: $$\frac12bl\sin \alpha+\frac12cl\sin \alpha=\frac12bc\sin 2\alpha \Rightarrow \\ l=\frac{2bc\cos \alpha}{b+c}\Rightarrow \\ l^2=\frac{4b^2c^2\cos ^2\alpha}{(b+c)^2}$$ Apply Cosine theorem to $\Delta ABC$: $$a^2=b^2+c^2-2bc\cos 2\alpha \Rightarrow \\ \cos 2\alpha=2\cos^2\alpha-1=\frac{b^2+c^2-a^2}{2bc} \Rightarrow \\ \cos^2\alpha=\frac{(b+c)^2-a^2}{4bc}$$ Hence: $$l^2=\frac{4b^2c^2\cos ^2\alpha}{(b+c)^2}=\frac{4b^2c^2}{(b+c)^2}\cdot \frac{(b+c)^2-a^2}{4bc}=bc\left(1-\frac{a^2}{(b+c)^2}\right).$$ as indicated by Lozenges in the comment.

$2.$ Internal angle bisector theorem: $$\frac{AC}{CU}=\frac{AB}{BU} \Rightarrow \frac{b}{CU}=\frac{c}{a-CU} \Rightarrow CU=\frac{ab}{b+c}$$ External angle bisector theorem: $$\frac{BU'}{CU'}=\frac{AB}{AC} \left(=\frac{AB}{AD}\right) \Rightarrow \frac{a+CU'}{CU'}=\frac{c}{b} \Rightarrow CU'=\frac{ab}{c-b}$$ From the right angle triangle $AUU'$: $$AU'^2=UU'^2-AU^2=(CU'+CU)^2-bc\left(1-\frac{a^2}{(b+c)^2}\right)=\\ \left(\frac{ab}{c-b}+\frac{ab}{b+c}\right)^2-bc\left(\frac{(b+c)^2-a^2}{(b+c)^2}\right)=\\ \vdots\\ =bc\left(\frac{a^2}{(c-b) ^2}-1\right)$$ Note: Details (e.g. working with angles, triple dot, etc) are left out to

1) keep it short and simple,

2) allow OP to practice.

Also, see algebraic simplification for the last step: WA answer.

Answer 2

You will need three equations: $$AU^2=c^2+BU^2-2cBU\cos(\beta)$$ $$\cos(\beta)=\frac{a^2+c^2-b^2}{2bc}$$ $$BU:(a-BU)=c:b$$ $$AU^2=c^2+BU^2-BU\left(\frac{a^2+c^2-b^2}{b}\right)$$ $$BU=\frac{ac}{b+c}$$ It is $$AU^2=c^2+\left(\frac{ac}{b+c}\right)^2-\frac{ac}{b+c}\left(\frac{a^2+c^2-b^2}{b}\right)$$