Without loss of generality,we may assume that $D$ is inside $\triangle PAB$ and $E$ outside $\triangle PAB$. Since $\angle AOD=\angle COD,\angle BOE=\angle COE$ and $OA=OB=OC$,we obtain $\triangle AOD\cong \triangle COD$ and $\triangle BOE\cong \triangle COE$. Hence $CD=AD,CE=BE,\angle OAD=\angle OCD,\angle OBE=\angle OCE$. Moreover,$\angle PAD=90^\circ-\angle OAD=90^\circ-\angle OCD=90^\circ-(180^\circ-\angle OCE)=90^\circ-(180^\circ-$ $\angle OBE)$ $=\angle OBE-$ $90^\circ=\angle PBE$. Note that $PA^2+AD^2-PD^2=PA^2+CD^2-PD^2=PA^2-PC^2$. Analogously,$PB^2+BE^2-PE^2=PB^2+CE^2-PE^2=PB^2-PC^2$. Combining $\cos \angle PAD=\cos \angle PBE$,we get that $2PA\cdot AD=2PB\cdot BE$,which follows that $AD=BE$.Consequently,$CD=CE$,as desired. $Q.E.D.$

Let $\gamma$ denote the tangent of $(O)$ at $C.$ Since $CP$ is C-symmedian of $\triangle ABC,$ the pencil $CA,CB,CP,\gamma$ is harmonic. $OD,OE$ are perpendicular bisectors of $\overline{CA},\overline{CB}$ $\Longrightarrow$ pencil $OD,OE,\tau,OC$ formed by perpendiculars from $O$ to $CA,CB,CP,\gamma$ is also harmonic $\Longrightarrow$ $D,E,C$ and $\tau \cap DE$ (at infinity) are harmonically separeted $\Longrightarrow$ $\overline{CD}=-\overline{CE}.$

See attachment for my solution

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@XmL Here is a similar way to deal with the segments based on your solution. $\angle CAM=\frac{1}{2}\angle COB=\angle COE$,which follows that $\triangle AMC\sim \triangle OCE$.Hence $\frac{CM}{CE}=\frac{AM}{OC}$. Analogously,$\triangle BMC\sim \triangle OCD$ and $\frac{CM}{CD}=\frac{BM}{OC}$.Consequently,since $AM=BM$,we obtain $CD=CE$.

liuyj8526 wrote: Let $PA, PB$ be tangents to a circle centered at $O$, and $C$ a point on the minor arc $AB$. The perpendicular from $C$ to $PC$ intersects internal angle bisectors of $AOC,BOC$ at $D,E$. Show that $CD=CE$ Actually this question is from China Western Mathematical Olympiad, which was renamed as China Western Mathematical Invitational Cmpetition (CWMI) since 2012...

Let $\Gamma$ be the circle centered at $O$ through $A,B$ and let $X$ be the second intersection of $PC$ and $\Gamma.$ The case $AC=AB$ is trivial, so assume WLOG that $AC<AB.$ Let $OD\cap AC=M.$ Then $\angle ODC=\angle MDC.$ Since $XC\perp DC$ and $OD\perp AC,$ we easily get $\angle MDC=\angle XCA$ so $\angle ODC=\angle XCA.$ Furthermore, $\angle DOC=\dfrac{\overarc{AC}}{2}=\angle AXC$ so by AA we may conclude $\triangle XCA\sim \triangle ODC,$ and similarly $\triangle XCB\sim \triangle OEC.$ Thus $\dfrac{XB}{CB}=\dfrac{OC}{EC},\dfrac{XA}{CA}=\dfrac{OC}{DC}\implies \dfrac{DC}{EC}=\dfrac{XB}{XA}\cdot \dfrac{CA}{CB}=1,$ where the last equality follows from $ACBX$ harmonic (follows from definition of $X,$ since $XC\cap AA\cap BB=P$). Thus $DC=EC$ and we may conclude.

What does the perpendicular from C to PC means??

Another idea, it seems: Let $PC$ interssect $O$ again at $F$; $AFBC$ is a harmonic quadrilateral and, if $Q$ is midpoint of $AB, \widehat{BFQ}=\widehat{AFC}$. Let's assume $D$ on the bisector of $\angle AOC$. We have $OD\bot AC, DE\bot CF$ and $\angle ODC=\angle ACF$, but $\angle ACF=\angle ABF$, also $\angle AFB=\frac{\angle AOB}2=\angle DOE$, thus $\triangle DOE\sim\triangle BFD$. As $\angle DOC=\angle AFC=\angle BFQ, OC$ and $FQ$ are homologous lines, thus $C$ is midpoint of $DE$, done. See drawing, same color angles are equal. Best regards, sunken rock

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