Geometry - Incircle I of $\triangle$ABC with chord MN intersecting AB at P. If BP = MC, find $\angle$AIC I recently found the following problem:

Circle with center $I$ is inscribed in triangle $ABC$ and touches the sides $A$C and $BC$ in points $M$ and $N$. The line $MN$ intersect the line $AB$ at $P$, as $B$ is between $A$ and $P$. If $BP = CM$ , find $\angle AIC$, in degrees.

This is from IWYMIC 2011
 A: 
Let the side lengths of the triangle ABC be $a$, $b$ and $c$. Apply the sine rule to the triangles PBN and AIC,
$$\frac{\sin\alpha}{\sin\beta} = \frac{BN}{BP} = \frac{BN}{MC}
=\frac {\frac12 (a+c-b)}{{\frac12 (a+b-c)}} =\frac {a+c-b}{{a+b-c}}\tag{1}$$
$$\frac{\sin\alpha}{\sin(180-\beta)} = \frac{AM}{AP}= \frac{AM}{AB+MC}=\frac {\frac12 (b+c-a)}{{c+\frac12 (a+b-c)}} = \frac {b+c-a}{{a+b+c}}\tag{2}$$
where the incircle results, 
$$BN = \frac 12 (a+c-b);\>\>\>MC = \frac 12 (a+b-c);\>\>\>
AM = \frac 12 (c+b-a)$$
are used. Combine (1) and (2),
$$\frac {a+c-b}{{a+b-c}} = \frac {b+c-a}{{a+b+c}}$$
which leads to $a^2+c^2=b^2$. Thus, ABC is a right triangle, with $\angle B = 90$. So,
$$\angle AIC = 180 - \frac{\angle A+\angle C}{2} = 180 - \frac 12 \angle B = 135^\circ$$
A: Following rather standard notations, let $a,b,c$ the be lengths of the sides in $\Delta ABC$, let $p=(a+b+c)/2$ be its half-perimeter.
Applying the theorem of Menelaus in $\Delta ABC$ w.r.t. the line $PMN$, we have
$$
1
=
\frac{\color{gray}{PB}}{PA} \cdot
\frac{MA}{\color{gray}{MC}} \cdot
\frac{NC}{NB} 
=
\frac{MA}{PA} \cdot
\frac{NC}{NB} 
=
\frac{p-a}{c+(p-c))}\cdot
\frac{p-c}{p-b}\ .
$$
This implies:
$$
\begin{aligned}
0 &= 2p(p-b)- 2(p-a)(p-c) = 2p(a+c-b)-2ac\\
&=((a+c)+b)((a+c-b))-2ac =a^2+c^2-b^2\ .
\end{aligned}
$$
So the given triangle has a right angle in $B$,
which then gives $\frac 12(\hat A+\hat C)=\frac 12\cdot 90^\circ =45^\circ$,
so $\widehat{AIC}=180^\circ -45^\circ =135^\circ$.

Later edit: Picture was inserted for an easy compilation of the above:

A: 
A  geometric solution:
Mark the tangent point of AB and circle as E. The quadrilateral EBNI is square because we have:
$BE=BN$
$IE=IN$
and $\angle BNI=\angle BEI=90^o$
Therefore $\angle EBN(or ABC)=90^o$ 
Now IC  and IA are the bisctors of angles $\angle BCA$ and $\angle CAB$ (of triangle ABC) respectively and we can write:
$\angle AIC= 90+\frac{\angle ABC}{2}=90+\frac{90}{2}=135^o $
