Solution:
First we show that $\angle DEC = 30^{\circ}$. Choose a point $F$ on $AB$ such that $CF = CB$. Join $FC$ and $FD$. Observe that $\angle DCB = 75^{\circ} + 25^{\circ} = 100^{\circ}$. Since $CD = CB$, we have $\angle CDB = \angle CBD = 40^{\circ}$. Therefore $\angle CBF = 40^{\circ} + 30^{\circ} = 70^{\circ}$. This gives $\angle CFB = 70^{\circ}$.
Since $CD = CB = CF$, we have the isosceles triangle $CDF$. But $\angle BCF = 40^{\circ}$. Hence $\angle FCD = 60^{\circ}$. Therefore we have an equilateral triangle $CDF$. This means $FD = FC = CD$ and $\angle DFC = 60^{\circ}$.

Observe that $\angle AFC = 110^{\circ}$ and $\angle FCA = 35^{\circ}$. Hence $\angle FAC = 35^{\circ}$. This means $FA = FC = FD$. Thus $F$ is the circumcentre of $\triangle ADC$. This implies that
$$\angle CAD = \frac{\angle CFD}{2} = 30^{\circ}$$
Therefore $\angle DEC = \angle DAC = 30^{\circ}$. Now concentrate on triangle $DCE$.
Construct an equilateral triangle $ECG$ with $CE$ as base, on the side of $B$. Join $GD$.

We have $\angle CGE = \angle GCE = \angle CEG = 60^{\circ}$ and $CE = EG = GC$. Since $\angle CED = 30^{\circ}$, we get $\angle GED = 30^{\circ}$. Thus $ED$ is the angle bisector of the isosceles triangle $GEC$. This implies that $ED$ is also the perpendicular bisector of $GC$. Thus $D$ is on the perpendicular bisector of $GC$. Therefore $DC = DG$ and hence $\angle DGC = \angle DCG$.
But $\angle DCG = 100^{\circ} - 60^{\circ} = 40^{\circ}$. This implies that $\angle DGC = 40^{\circ}$ and hence $\angle CDG = 100^{\circ}$.
Consider the quadrilateral $GBCD$. We have $DG = DC = CB$, $\angle GDC = 100^{\circ} = \angle DCB$. It is an isosceles trapezium. (or we can show that $\triangle GDC \cong \triangle BCD$.) Therefore $DB = GC$. But $GC = CE$. Thus we get $DB = CE$.

Alternate Solution
As in the previous solution, one shows that $F$ is the circumcenter of $\triangle ADC$. Since $E$ lies on this circumcircle, this means $FE$ is equal to all of the sides $FA, FD, FC$ and thus also to $CD$ and $CB$. Now $CDB$ and $FCE$ are both isosceles triangles with base angles $40^{\circ}$, and they have $CD = FC$, so they are in fact congruent. This directly implies $CE = BD$, as required.