Die This is Exercise 3-5 from the Art of Problem Solving Volume 2 by Richard Rusczyk and Sandor Lehoczky. I looked at the solution in the solution manual, but I don't quite understand it, so I'm hoping that someone here could explain how to solve this problem
In triangle $ABC$, let $D$ be a point on $BC$ such that $AD$ bisects angle $A$. If $AD=6$, $BD=4$, and $DC=3$, then find $AD$.
The solution they gave is as follows:
Let $AB=x$ and $AC=y$. Since $\angle ADB = 180 - \angle ADC$, $\cos \angle ADB = -\cos \angle ADC$, and:
$$ (x^2-52)/-48 = -(y^2-45)/-36 $$
Rearranging this we find $3 x^2+4 y^2=336$. From the Angle Bisector Theorem we have $x/y = 4/3$, and our equation becomes $21 x^2/4 = 336$, so $x=8$.
My main issue with this is that I don't understand where they got the second line from.
*Edit: Sorry I may not have been clear enough before but when I said second line I meant the equation with the (x^2)-52/-48 and all that. I have no idea where those numbers came from and why x and y are squared, etc.
Remember that cosine is an even function, so $\cos (-\theta) = \cos \theta$. Also remember that the graph of a function $f(x-a)$, is just $f(x)$ shifted $a$ units to the right.
So, given two angles, $\alpha$ and $\beta$, if $\alpha = 180^\circ - \beta$, then $$\begin{align} \cos \alpha &= \cos(180^\circ - \beta) \\ &= \cos(\beta - 180^\circ) \\ &= -\cos \beta.\end{align}$$
If you're referring to why $x/y = 4/3$, then see the proof of the angle bisector theorem. It uses the law of sines, so it's fairly quick.