Length of line segment in a triangle

Question

Triangle ABC has $\vert AB\vert$= 2007 and $\vert AC \vert$ = 2015. The incircle of the triangle is tangent to AC and AB at E and F respectively, and P is the intersection point of EF and BC. Suppose B is the midpoint of CP. Compute the length $\vert BC \vert$.

Answer 1

First of all what is $|AD|$? If you meant $|AB|$, apply Menelaus' theorem and then form two very simple equations to find $|BF|$. If you don't know Menelaus' theorem, simply draw a line through $B$ parallel to $EF$.

Answer 2

So assuming $B=D$ you set the following notations:

$G$ the tangent point of the incircle with $BC$, $x=|BC|$ and $$|AE|=|AF|=a\\ |BF|=|BG|=b\\ |CG|=|CE|=c$$

By Menelaüs theorem: $$\frac{PB}{PC}\frac{EC}{EA}\frac{FA}{FB}=1$$ Thus you get: $\frac{1}{2}\frac{c}{a}\frac{a}{b}=1$ so $c=2 b$.

Now you have the following equations: \begin{align} a+c &=2015\\ a+b &=2007\\ b+c &=x \end{align} Thus $b=c-b=2015-2007=8$ and $x=3 b =24$

Conclusion: $|BC|=24$.