Prove that $MP//CQ$. Let the triangle $ABC$ and the incircle $(I)$. $M, N$ are in $AB, AC$ resp such that $MN//BC$. Let $(K)$ be the excircle of the triangle $AMN$ respect with $A$. $MP, CQ$ are the tangent lines of $(I), (K)$ resp ( $MP$ differents $AB$, $CQ$ differents $AC$). Prove that $MP//CQ$.
I think one can prove $IP//KQ$, but i do not know how to do it. Can anyone help me?
 A: 
Let $|BC|=a$, $|CA|=b$, $|AB|=c$,
$\angle BAC=\alpha$,
$\angle ABC=\beta$,
$\angle BCA=\gamma$,
$S$ is the area of $\triangle ABC$,
$\rho=\tfrac12(a+b+c)$ is its semiperimeter,
$|IP|=|IC|=r$ is its inradius 
and $r_a=\rho\,\tan\tfrac\alpha2$ 
is exradius of $\triangle ABC$ with respect to $A$.
Let $|AM|=u\,c$ for some $u\in(0,1)$, 
that is, $\triangle AMN$ is a scaled down version of $\triangle ABC$.
Then the exradius of $\triangle AMN$ with respect to $A$
is $|KQ|=|KQ_2|=|KK_c|=u\,r_a$.
Let's check that $\angle PMB=\angle CQ_1B$.
From $\triangle IPM$ and $\triangle IMI_c$ we have
\begin{align}
 \angle BMP&=
 2\,\angle BMI
 =2\,\arctan\frac{r}{|MI_c|}
 =2\,\arctan\frac{r}{|AI_c|-u\,c}
 =2\,\arctan\frac{r}{r\,\cot\tfrac\alpha2-u\,c}
,
\end{align}
and from $\triangle AQ_1C$,
$\triangle CKQ$,
$\triangle CQ_2K$
and $\triangle BQ_1C$, we have
\begin{align}
\angle CQ_1B&
=\alpha +2\,\angle QCK
=\alpha +2\,\angle KCQ_2
=\alpha+2\,\arctan\frac{u\,r_a}{b-u\,r_a\,\cot\tfrac\alpha2}
.
\end{align}
Now let's check that $\tan(\tfrac12\angle PMB)=\tan(\tfrac12\angle CQ_1B)$.
\begin{align}
\tan(\tfrac12\angle PMB)
&=
\frac{r}{r\,\cot\tfrac\alpha2-u\,c}
=
\frac{r\tan\tfrac\alpha2}{r-u\,c\,\tan\tfrac\alpha2}
=
\frac{S\tan\tfrac\alpha2}{S-u\,c\,\rho\tan\tfrac\alpha2}
=
\frac{S\tan\tfrac\alpha2}{S-u\,c\,r_a}
\tag{1}\label{1}
,\\
\tan(\tfrac12\angle CQ_1B)
&=
\tan\left(
\tfrac\alpha2+\arctan\frac{u\,r_a}{b-u\,r_a\,\cot\tfrac\alpha2}
\right)
\\
&=
\frac{b\,\tan\tfrac\alpha2}{b-u\,r_a\,(\cot\tfrac\alpha2+\tan\tfrac\alpha2)}
=
\frac{b\,\tan\tfrac\alpha2}{b-u\,r_a\,\frac{2}{\sin\alpha}}
=
\frac{\tfrac12\,b\,c\,\sin\alpha\tan\tfrac\alpha2}{\tfrac12\,b\,c\,\sin\alpha-u\,c\,r_a}
\\
&=
\frac{S\,\tan\tfrac\alpha2}{S-u\,c\,r_a}
\tag{2}\label{2}
.
\end{align}
\eqref{1}=\eqref{2}, hence $\angle PMB=\angle CQ_1B$
and $MP || CQ$.
