Geometry problem on the incentre and circumcenter of a triangle I have the following problem:

In a triangle $ABC$ the line joining incentre and circumcentre is parallel to side $BC$. Prove that $\cos B + \cos C=1$.

Could someone help me solve it?

This one has been untouched for a few days now, so here's a solution. It does the job but it feels like I beat it into submission with a stick.
I would be interested to know if anyone has a more elegant approach, as it was tougher than I was expecting. 

In the above diagram $I$ is the incentre and $K$ is the circumcentre of $\Delta$ ABC.
$PS \parallel BC $ therefore
$$ \frac{CP}{BS}=\frac{AC}{AB}= \frac{ \sin B}{\sin C}, \quad (1)$$
by the sine rule.
Furthermore, $\Delta$ CIP and $\Delta$ BIS are isosceles and so
$$ \frac{PS}{BC}=\frac{CP+BS}{BC}= 1 - \frac{BS}{AB},$$
the last equality is implied by $PS \parallel BC .$ And so
$$ CP = \left( 1 - \frac{BS}{AB} \right) BC - BS = \frac{\sin B}{\sin C} \cdot BS,$$
using $(1).$
Hence
$$ BC = BS \left( 1+ \frac{\sin B}{\sin C} + \frac{BC}{AB} \right)
= BS \left( 1+ \frac{\sin B}{\sin C} + \frac{\sin (B+C)}{\sin C} \right), \quad (2)$$
by the sine rule.
Let $ \angle CBK=\theta$ then, again by the sine rule and noting that $BK=CK,$
we have
$$\frac{\sin B}{\sin (B+C)} = \frac{AC}{BC} = \frac{\cos (C-\theta)}{\cos \theta} =
\cos C + \sin C \tan \theta.$$
Hence
$$ \tan \theta = \frac{ \frac{\sin B}{\sin (B+C)} - \cos C}{\sin C}. $$
Also
$$BC = 2 RK \cot \theta = 2 BS \sin B \cot \theta
= \frac{2BS \sin B \sin C}{ \frac{\sin B}{\sin(B+C)} - \cos C}, \quad (3)$$
using $RK = BS \sin B$ and the expression for $\tan \theta.$
Combining $(2)$ and $(3)$ we obtain
$$ 1+ \frac{\sin B}{\sin C} + \frac{\sin (B+C)}{\sin C}
=\frac{2\sin B \sin C}{ \frac{\sin B}{\sin(B+C)} - \cos C}. \quad (4)$$
Now we need to show that $(4) \Rightarrow \cos B + \cos C = 1.$
ABC is a non-degenerate triangle and so $C \ne 0$ and $\cos B + \cos C \ne -1$ and so
$\cos B + \cos C = 1 \Longleftrightarrow$
$$2\sin B \sin^2 C = 2 \sin B (1- \cos^2 C) + \sin C
\lbrace 1 - (\cos B + \cos C)^2 \rbrace $$
$$= 2\sin B - 2\sin B \cos^2 C + \sin C (-2\cos B \cos C + \sin^2 B - \cos^2 C)$$
$$=\lbrace 2 + \cos(B-C) \rbrace \lbrace \sin B - \cos C \sin(B+C) \rbrace, \quad (5)$$
noting that $\sin(B+C)\cos(B-C) = \sin B \cos B + \sin C \cos C.$ When we expand the RHS of $(5)$ the two terms $\sin B \cos B \cos C$ and $-\sin B \cos B \cos C$ cancel, leaving precisely the RHS of the second to last line of $(5).$
Hence we have $\cos B + \cos C = 1 \Longleftrightarrow$
$$ \frac{2+ \cos(B-C)}{\sin C} = \frac{2 \sin B \sin C}{\sin B - \cos C \sin (B+C)}. \quad (6)$$
Now
$$\sin (B+C) \lbrace 1+ \cos(B-C) \rbrace = \sin(B+C) + \sin(B+C) \cos(B-C)$$
$$=\sin B \cos C + \cos B\sin C + \sin B \cos B + \sin C \cos C $$
$$=(\sin B + \sin C)(\cos B + \cos C) = \sin B + \sin C \quad (7)$$
$\Longleftrightarrow \cos B + \cos C = 1,$ since $\sin B + \sin C \ne 0.$
And so the LHS of $(4)$
$$ 1+ \frac{\sin B}{\sin C} + \frac{\sin (B+C)}{\sin C} =
\frac{\sin B + \sin C + \sin(B+C)}{\sin C}$$
and using $(7)$ to substitute for $\sin B + \sin C$
$$=\frac{2\sin(B+C)+ \sin(B+C)\cos(B-C)}{\sin C}
=\sin(B+C)\frac{2+\cos(B-C)}{\sin C}$$
and using $(6)$ to substitute for the fraction
$$=\frac{2\sin B \sin C \sin(B+C)}{\sin B - \cos C \sin(B+C)}
=\frac{2\sin B \sin C}{ \frac{\sin B}{\sin(B+C)} - \cos C}.$$
And so we have shown that when $B$ and $C$ are angles of a triangle then
$(4) \Rightarrow \cos B + \cos C = 1$ and the result is proved.
