1. Define the Simson Line and Euler Line:
- The Simson line of a point $P$ with respect to a triangle $\triangle ABC$ is the line passing through the feet of the perpendiculars dropped from $P$ to the sides of $\triangle ABC$.
- The Euler line of a triangle is the line that passes through several important points of the triangle, including the orthocenter $H$, the centroid $G$, and the circumcenter $O$.

2. Given Conditions:
- $ABCD$ is a cyclic quadrilateral.
- The Simson line of $A$ with respect to $\triangle BCD$ is perpendicular to the Euler line of $\triangle BCD$.

3. Construct Points and Lines:
- Let $E$ and $F$ be the feet of the perpendiculars from $B$ to $AC$ and $DC$, respectively.
- Let $B^*$ be the intersection of the altitude from $B$ to $DC$ with the circumcircle of $ABCD$.
- Similarly, define $A^*$ as the intersection of the altitude from $A$ to $BD$ with the circumcircle of $ABCD$.
- Let $M$ be the midpoint of $AB^*$ and $M^*$ be the midpoint of $A^*B$.
- Let $H_1$ and $H_2$ be the orthocenters of $\triangle ADC$ and $\triangle BDC$, respectively.

4. Lemma:
- The Simson line of $B$ with respect to $\triangle ACD$ is parallel to $AB^*$.

5. Proof of Lemma:
- Consider the angles:
$$\angle BB^*A = \angle BCA = \angle BFE$$
Since $\angle BCA = \angle BFE$, it follows that $EF \parallel AB^*$.

6. **Simson Line of $A$:**
- Similarly, the Simson line of $A$ with respect to $\triangle BCD$ is parallel to $A^*B$.

7. Collinearity and Cyclic Quadrilateral:
- Note that $M, O, H_1$ are collinear.
- The triangle $\triangle AH_1B^*$ is isosceles, so:
$$\angle H_1AB^* = \angle HB^*A$$
- Since $H_1H_2 \parallel AB$, the quadrilateral $H_1H_2B^*A^*$ is cyclic.

8. Angle Relationships:
- We know:
$$\angle AA^*B = \angle AB^*B$$
Therefore:
$$\angle H_1B^*A = \angle H_2A^*B$$
- This implies:
$$\angle A^*AB^* = \angle A^*BB^*$$
and:
$$\angle AB^*H = \angle H_2A^*B$$
- Thus, the triangle $\triangle BH_2A^*$ is isosceles.

9. Collinearity of Points:
- Points $O, H_1, H_2, M^*, M$ are collinear.

10. Conclusion:
- Since the Simson line of $A$ with respect to $\triangle BCD$ is perpendicular to the Euler line of $\triangle BCD$, by symmetry and the properties of the cyclic quadrilateral, the Simson line of $B$ with respect to $\triangle ACD$ is also perpendicular to the Euler line of $\triangle ACD$.

$\blacksquare$