<s>[INST] 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
 [/INST] 
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,
$$\angle AIC = 180 - \frac{\angle A+\angle C}{2} = 180 - \frac 12 \angle B = 135^\circ$$
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