Identify the common angle.

Both triangles $\triangle BOC$ and $\triangle BED$ share the angle $\angle B$.

$$\angle B = \angle B \quad (\text{Common angle})$$

From the previous problem (3.1.1), we determined that $\hat{A} = y$.
By the Tangent-Chord Theorem, the angle between the tangent BDC and the chord DE ($\angle BDE$ or $\hat{D}_4$) is equal to the angle in the alternate segment ($\hat{A}$).

$$\angle BDE = \hat{A} = y$$

Also from 3.1.1, given that $CO \parallel DE$, the alternate interior angles formed by transversal CE are equal.

$$\angle OCE = \angle DEC$$ $$\hat{C}_1 = \hat{E}_1 = y$$

Therefore, we have:

$$\angle C_1 = y$$

Comparing $\angle C_1$ from $\triangle BOC$ and $\angle BDE$ from $\triangle BED$:

$$\angle C_1 = \angle BDE = y$$

Since two angles of $\triangle BOC$ are equal to two corresponding angles of $\triangle BED$ ($\angle B = \angle B$ and $\angle C_1 = \angle BDE$), the triangles are similar by the Angle-Angle (AA) similarity criterion.
Thus, we can write the similarity correspondence as:

$$\triangle BOC \sim \triangle BED$$