<s>[INST] A triangle $ABC$ with $c=60,l_c=10$ and $P=125$ A triangle $ABC$ is given with $c=60,l_c=10$ and $P=125$, where $AB=c$, $l_c$ is the angle bisector of $\measuredangle C$ and $P$ is the perimeter. Find $a$ and $b$.

Here are my thoughts:
We have $\begin{cases}\dfrac{m}{n}=\dfrac{b}{a}\\l_c^2=ab-mn\end{cases}.$ From the first equation we can get $$\dfrac{m+n}{n}=\dfrac{a+b}{a}$$ or $$\dfrac{12}{n}=\dfrac{13}{a}.$$ Then we can substitute $l_c=10$ into the second equation to get $$100=ab-mn.$$ I am stuck here. Thank you in advance!
Katherine
 [/INST] Length of angle bisector is given by
$\displaystyle l_c^2 = \frac{ab}{(a+b)^2} ((a+b)^2 - c^2)$
$a+b = 125-60 = 65, c = 60$
Leads to $\displaystyle 10^2 = \frac{ab}{65^2}(65^2-60^2)$
i.e., $ab = 676 \implies a(65-a) = 676$
We get a quadratic $a^2 - 65a + 676 = 0 \implies a = 13, 52$
So $a = 13, b = 52$ or $a = 52, b = 13$
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