Let $ABC$ be a triangle in which $\angle ABC = 45^o$ and $\angle BAC > 90^o$. Let $O$ be the midpoint of the side $[BC]$. Consider the point $M \in (AC)$ such that $\angle COM =\angle CAB$. Perpendicular from $M$ on $AC$ intersects line $AB$ at point $P$. a) Find the measure of the angle $\angle BCP$. b) Show that if $\angle BAC = 105^o$, then $PB = 2MO$.