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 $\angle LOI =\angle CPI$.
 A: 
Let $N$ be the midpoint of $CD$ and let $M$ be the midpoint of $AB$. Note that $ONLM$, having two right angles, is cyclic. Using the power of a point theorem, we now find
$$
DC \cdot DL = 2 \cdot DN \cdot DL = 2 \cdot DO \cdot DM = DO \cdot DP,
$$
implying that $C$, $L$, $O$ and $P$ are concyclic.
It now suffices that $\angle LIO = \angle IPO$. Indeed, if $\angle LIO = \angle IPO$ then
we find $$\begin{align*}
\angle CPI &= \angle CPO - \angle IPO \\&= \angle CLO - \angle IPO \\&= \angle CLO - \angle LIO \\&= \angle ILO - \angle LIO \\&= - \angle OLI - \angle LIO = \angle IOL
\end{align*}
$$ (we are working with directed angles modulo $180^\circ$ here).
In other words, it remains to show that $CI$ is tangent to the circumcircle of $\triangle IPO$. However, this follows immediately from $DO \cdot DP = DL \cdot DC = DI^2$, where the last equality follows from $DI^2 = DB^2 = DL \cdot DC$ ($DB$ is tangent to the circumcircle of $\triangle BLC$ and $\triangle DIB$ is isosceles).
A: 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.
