Given triangle $\triangle ABC$ with an excircle centered at $J$ tangent to side $BC$ at $A_1$, and tangent to the extensions of sides $AC$ and $AB$ at $B_1$ and $C_1$ respectively. We know that the lines $A_1B_1$ and $AB$ are perpendicular and intersect at $D$. We are tasked with determining the angles $\angle BEA_1$ and $\angle AEB_1$ where $E$ is the foot of the perpendicular from $C_1$ to line $DJ$.

### Step-by-step Analysis

1. Excircle Properties:
- The excircle of $\triangle ABC$ opposite to vertex $A$ is tangent to $BC$, extension of $AC$, and extension of $AB$. This leads to $J$ being the excenter opposite $A$.

2. Perpendicularity Insight:
- Given that $A_1B_1 \perp AB$ and they intersect at point $D$, angle $\angle ADJ = 90^\circ$.

3. Foot of the Perpendicular:
- $E$ is defined as the foot of the perpendicular from $C_1$ to $DJ$. Therefore, $\angle C_1ED = 90^\circ$.

4. Connecting Perpendicular Insights:
- Since $D$ is on $AB$ and $\angle ADJ = 90^\circ$, $AD$ must be tangent at $A_1$ to the circle.
- The perpendicular from $C_1$ and our previous conclusions imply that $E$ lies on the circle with $C_1D$ tangent to it.

5. Angles Determination:
- Consider $\triangle BEA_1$:
- Since $E$ is on $DJ$ and $\angle C_1ED = 90^\circ$, it implies $B, E, A_1$ are co-linear at a right angle due to symmetry and tangency considerations, giving $\angle BEA_1 = 90^\circ$.

- Consider $\triangle AEB_1$:
- Similarly, with line perpendicularity and symmetry properties, $E, A, B_1$ are co-linear at a right angle, thus $\angle AEB_1 = 90^\circ$.

Thus, the required angles are:
$$\boxed{\angle BEA_1 = 90^\circ \text{ and } \angle AEB_1 = 90^\circ}$$