1. **Problem Statement:** We have a cyclic quadrilateral ABCD inscribed in a circle with diagonals AC and BD intersecting at point E. Given angles $\angle A = 25^\circ$ and $\angle B = 75^\circ$, we need to find the angle $x = \angle AED$ at the intersection point E of the diagonals. 2. **Key Properties:** - In a cyclic quadrilateral, opposite angles sum to 180°: $\angle A + \angle C = 180^\circ$ and $\angle B + \angle D = 180^\circ$. - The intersection of diagonals in a cyclic quadrilateral creates vertical angles and related angle pairs. 3. **Calculate missing angles:** - Since $\angle A = 25^\circ$, then $\angle C = 180^\circ - 25^\circ = 155^\circ$. - Since $\angle B = 75^\circ$, then $\angle D = 180^\circ - 75^\circ = 105^\circ$. 4. **Using the intersecting chords theorem:** - The angle formed by two intersecting chords is half the sum of the measures of the arcs intercepted by the angle and its vertical angle. - Specifically, $\angle AED = \frac{1}{2}(\angle A + \angle C)$. 5. **Calculate $x = \angle AED$:** $$ x = \frac{1}{2}(25^\circ + 155^\circ) = \frac{1}{2}(180^\circ) = 90^\circ $$ 6. **Conclusion:** The angle $x$ at point E where the diagonals intersect is $90^\circ$.