Proof of the Angle Bisector Theorem in Triangle ABC

Consider a triangle ABC with the angle bisector of angle A intersecting side BC at point U. Our goal is to prove that $AU^2 frac{bc(1-a^2)}{b c}$. To achieve this, we will employ several fundamental principles from triangle geometry, trigonometry, and algebra. Let's proceed step-by-step.

Step 1: Setting Up the Problem

Firstly, let's introduce the given conditions and notations:

Let $xy a$. Given that $frac{x}{b} frac{y}{c} Rightarrow y frac{cx}{b}$, we can express $x$ in terms of $a$ and $c$: $x frac{ab}{b c}$. Similarly, we can find $y$: $y frac{ac}{b c}$.

Step 2: Using Trigonometric Identities

We will now use trigonometric identities to express cosine $theta$ in two different ways. On the right side:

$cos theta frac{d^2 - y^2 - c^2}{2dy}$

On the left side:

$cos theta -frac{d^2 - x^2 - b^2}{2dx}$

Step 3: Setting the Equations Equal

Since both expressions represent the same angle $theta$, we can set them equal to each other:

$frac{d^2 - y^2 - c^2}{2dy} -frac{d^2 - x^2 - b^2}{2dx}$

Let's simplify this equation step-by-step:

Multiply both sides by $2dxy$ to clear the denominators: $(d^2 - y^2 - c^2)x -(d^2 - x^2 - b^2)y$ Expand and rearrange terms: $d^2x - y^2x - c^2x -d^2y x^2y b^2y$ Collect terms involving $x$ and $y$: $d^2x d^2y x^2y y^2x c^2x b^2y$ Substitute the expressions for $x$ and $y$ in terms of $a$, $b$, and $c$: $d^2 left(frac{ab}{b c}right) d^2 left(frac{ac}{b c}right) left(frac{ab}{b c}right)^2 left(frac{ac}{b c}right) left(frac{ac}{b c}right)^2 left(frac{ab}{b c}right) c^2 left(frac{ab}{b c}right) b^2 left(frac{ac}{b c}right)$ Simplify the equation: $d^2 left(frac{ab ac}{b c}right) frac{a^2b^2c a^2c^2b c^2ab^2 b^2ac^2}{(b c)^2} c^2 left(frac{ab}{b c}right) b^2 left(frac{ac}{b c}right)$ Further simplification gives us: $d^2 left(frac{a(b c)}{b c}right) frac{a^2bc(b c) c^2ab(b c) b^2ac(b c)}{(b c)^2}$ Finally, after simplification, we derive the desired result: $AU^2 frac{bc(1-a^2)}{b c}$

Conclusion

We have successfully proven that the squared length $AU^2$ in the given scenario is equal to $frac{bc(1-a^2)}{b c}$. This proof combines algebraic manipulation and trigonometric identities to establish a relationship between the sides and angles of the triangle, demonstrating the power of geometric and trigonometric theorems in solving complex problems.