Let $N$ be the midpoint of the arc $ABC$, and $T$ be the midpoint of the lesser arc $NC$. Prove that $N$ and $T$ lie on $A'B'$. For this purpose, notice that $AN = BN = AB = A'B = B'A$.

From the parallelism of the lines $AB'$ and $BC$, we get that $\angle AB'B = \angle CBB' = \angle ABB'$. Therefore, $AB' = AB$. Similarly, $AB = A'B$. Denote $\angle BAC = 2\alpha$, $\angle ABC = 2\beta$. Let, without loss of generality, $\alpha > \beta$.

Let $N$ be the midpoint of the arc $ACB$ of the circle $\Omega$ (see Fig. 12). Then $AN = BN$ and $\angle ANB = \angle ACB = 60^\circ$; therefore, $= B'A$. Therefore, point $A$ is the center of the circle described around $-\angle AB' = 60^\circ - \beta$, whence $\angle NAB' = 2\angle NBB' = 120^\circ - 2\beta$ and

$= AN = BN$ and $\angle ANB = \angle ACB = 60^\circ$; therefore, $= B'A$. Therefore, point $A$ is the center of the circle described around $-\angle AB' = 60^\circ - \beta$, whence $\angle NAB' = 2\angle NBB' = 120^\circ - 2\beta$ and

$\angle ANB' = 90^\circ - \angle NAB'/2 = 30^\circ + \beta$. Similarly, $\angle BNA' = 30^\circ + \alpha$, whence $\angle B'NA + \angle ANB + \angle BNA' = (30^\circ + \beta) + 60^\circ + (30^\circ + \alpha) = 120^\circ + (\alpha + \beta) = 180^\circ$. Thus, point $N$ lies on the line $A'B'$.

Let $T$ be the midpoint of the lesser arc $NC$ of the circle $\Omega$. Note that $\angle ANT = \angle ABT = (\angle ABN + \angle ABC)/2 = 30^\circ + \beta = \angle ANB'$. Therefore, point $T$ also lies on the line $A'B'$, and triangle $CDE$ coincides with triangle $CNT$. This triangle is isosceles, since $NT = TC$.