Given triangle $ABC$ with angle $C$ bisected to meet side $c$, we have the following sides of the triangle: $a$, $b$, $c$, $x$, $p$, and $q$ where $x$ is the angle bisector, $p+q=c$ (the two divisions created by the angle bisector.
We are expected to prove that $p=bc/(a+b)$ and (for the other triangle) $q=ac/(a+b)$. This likely uses LoS or LoC, but I'm not sure how.
On
You can think of $p$ and $q$ as solutions to the following system of quations
$$ \begin{cases} p+q = c \\\\ \dfrac{p}{b} = \dfrac{q}{a} \end{cases} $$
The easiest way is to solve for one variable in terms of the other and pug it into the other equation. For example, using $q= c-p$ we get
$$ \frac{p}{b} =\frac{c-p}{a} = \frac{c}{a}- \frac{p}{a} $$ $$ \frac{p}{a} + \frac{p}{b}= \frac{c}{a} $$ $$ p\left(\frac{1}{a}+\frac{1}{b} \right)= \frac{c}{a} $$ $$ p \cdot\frac{a+b}{ab}= \frac{c}{a} $$ $$ b = \frac{bc}{a+b} $$
Althernatively, you can also use $q = \dfrac{ap}{b}$ to get $$ p + \frac{ap}{b} = c $$ $$ p \left(1 + \frac{a}{b} \right) = c $$ $$ p \cdot \frac{a+b}{b}= c $$ $$ p = \frac{bc}{a+b} $$
Start off with the internal bisector theorem... $$\begin{align}\dfrac ba&=\dfrac pq\\\dfrac ab&=\dfrac qp\\\dfrac{a+b}b&=\dfrac{p+q}p\qquad(\text{Componendo})\\\dfrac{a+b}b&=\dfrac cp\\p&=\dfrac{bc}{a+b}\end{align}$$