Cho $\Delta ABC$, biết $BC=6$. Lấy $E, F$ theo thứ tự thuộc $AB, AC$ sao cho $EF$ song song với $BC$ và tiếp xúc với đường tròn nội tiếp $\Delta ABC$. Tính chu vi $\Delta ABC$, biết $EF=2$

Ta có hai $\Delta AEF$ và $\Delta ABC$ đồng dạng, do đó $\frac{EF}{BC}=\frac{AE}{AB}     (1)$
Ta có:
$\left\{ \begin{array}{l} c=AB=AM+MB=AM+NB\\ b=AC=AP+PC=AM+NC \end{array} \right. $
$\Rightarrow  b+c=2AM+BC=2AM=6$
$\Leftrightarrow  AM=AP=\frac{1}{2}(b+c-a)=p-a     (2) $
$\Rightarrow \left\{ \begin{array}{l} AM=AE+EM=AE+EQ\\ AM=AP=AF+FP=AF+FQ \end{array} \right.     \Rightarrow  2AM=AE+AF+EF$
$\Leftrightarrow  AM=AP=p_{\Delta AEF}     (3)$
Thay $(2), (3)$ vào $(1)$ ta được $\frac{2}{6}=\frac{p-6}{p}\Leftrightarrow  p=9  $

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