Let $CD$ and $BE$ intersect $AB$ and $AC$ at $F$ and $G$ respectively. Let $P$ be the intersection point of $CD$ and $BE$.

Because $\angle DEB = \angle DCB$ and $|DE| = |BC|$ it follows that $\triangle DPE$ is congruent to $\triangle BPC$. In particular, $|DP| = |BP|$. This implies $\angle BDP = \angle PBD$.

The quadrilateral $FPGA$ is cyclic as it has right angles at $F$ and $G$. Hence, $\angle BAC = \angle DPB$. Finally, $\angle BDP = \angle BDC = \angle BAC$ (both are subtended by $BC$), hence $\angle BDP$ is equilateral and so its internal angles are equal to $60^\circ$. Therefore, $\angle BAC = \angle DPB = 60^\circ$.

Let $CD$ and $BE$ intersect $AB$ and $AC$ at $F$ and $G$ respectively. Because of the right angles at $F$ and $G$ we have $\angle DAB = \angle DCB = 90^\circ - \angle ABC$ and $\angle CAE = \angle CBE = 90^\circ - \angle ACB$.

Therefore, the inscribed angle subtended by $DE$ is equal to
$$\begin{aligned} \angle DAE &= \angle DAB + \angle BAC + \angle CAE \\ &= 180^\circ - (\angle ABC + \angle ACB) + \angle BAC = 2\angle BAC. \end{aligned}$$

Because a chord that subtends (on either side) an inscribed angle $\alpha$ in a circle of radius $R$ has length $2R \sin \alpha$, we obtain now
$$\begin{aligned} |DE| &= 2R \sin (2\angle BAC) \\ &= 4R \sin (\angle BAC) \cos (\angle BAC) \quad \text{and} \\ |BC| &= 2R \sin (\angle BAC). \end{aligned}$$

From $|DE| = |BC|$ we get $\cos (\angle BAC) = 1/2$, thus $\angle BAC = 60^\circ$.