Bisector and symmetric point I can't solve the following problem and need help.
Given an acute-angled triangle ABC.CL is the bisector of $\angle C$ where $L \in AB$. CL instersects the circumcircle of the triangle at point D. O is the center of the circumcircle and I is the incenter of the triangle. If $P$ is the symmetric point of D with respect to $AB$, prove that $\angle LOI =\angle CPI$.

Since $\widehat{ACD}=\widehat{DCB}$, $D$ is the midpoint of the arc $AB$ in the circumcircle $\Gamma$ of $ABC$, hence $D,P,O$ are collinear and, obviously, $OC=OD$. 
Moreover $\widehat{DOC}=\widehat{BOC}+\widehat{BOD}=2\widehat{A}+\widehat{C}$, so $CD = 2R\sin(\widehat{A}+\widehat{C}/2) $. 

Now it is easy to check that:
$$ DP\cdot DO = DL\cdot DC, $$
hence $L,P,O,C$ lie on the same circle, so $\widehat{LPC}=\widehat{LOC}$ and we just need to prove that $\widehat{LPI}=\widehat{IOC}$. Since $DA=DB=DI$, we have that the circle $\Omega$ with center $D$ through $I$ is orthogonal to the circle through $L,P,O,C$, so:
 $$ DP\cdot DO = DL\cdot DC = DI^2 $$
gives that the circular inversion with respect to $\Omega$ switches $P,O$ and $L,C$, giving $\widehat{LPI}=\widehat{IOC}$ as wanted.
