Note that $\angle BDM = \angle BAC, \angle DMB = \angle BCA$, so $BDM \sim BAC$. So $\frac{BD}{AB} = \frac{BM}{BC}= \frac{CM}{BC}$, so $BD + CE = \frac{AB \cdot CM + AC \cdot BM}{BC} = \frac{AM \cdot BC}{BC} = AM$. Alternatively, note that $\angle DBA = \angle MBC = \angle MCB = \angle DAB$, so $DAB, EAC$ are $D, E$ isosceles. So $OD \perp AB, OE \perp AC$. Thus $OD, OE$ make equal angles with $AM$, so $D, E$ are isotomic wrt $AM$. Then we have $BD + CE = AD + AE = AD + DM = AM$.

Dear Mathlinkers, this is an application of http://www.artofproblemsolving.com/community/c6t48f6h1080717_easy_geometry noting also that the (BOM) and (COM) are equal...ODE is O-isoceles. Sincerely Jean-Louis

Let $\Gamma$ be circumscribed circle of triangle $ABC$ ($AB \ne AC$). Let O be circumcenter of triangle $ABC$. Let $M$ be a point where angle bisector of angle $\angle BAC$ intersects $\Gamma$. Let $D$ ($D\ne M$) be a point where circumscribed circle of triangle $BOM$ intersects line segment $AM$ and let $E$ ($E\ne M$) be a point where circumscribed circle of triangle $COM$ intersects line segment $AM$. Prove that $BD+CE=AM$.

Let $\angle{BCE}=\beta$ and $\angle{ECM}=\alpha$. $\implies \angle{BOM}=\angle{COM}=2\alpha+ 2\beta$. Since $DOMB$ and $COEM$ cyclic $\implies \angle{BDM}=\angle{CEM}= {2\alpha + 2\beta}$, and $\angle{BAM}=\angle{CAM}=\alpha+\beta$ this means $AD=DB$ and $AE=EC$.So $BD+CE=AD+AE$ $\implies$ we need to find $MD=AE$.From easy angle chasing we get $\triangle DBM \sim \triangle EMC$ and since $MB=MC$ we get $MD=CE=AE$ and we're done!