We are given the following information:

• $\angle BAC = 50^\circ$
• $\angle CAD = 60^\circ$
• $\angle CBD = 30^\circ$
• $\angle BDC = 25^\circ$
• $E$ is the point of intersection of diagonals $AC$ and $BD$.

We need to find $\angle AEB$.

Step 1: Use the Angle Sum Property in Triangle $ABC$

In triangle $ABC$, the sum of the angles is $180^\circ$. Therefore, $\angle ABC = 180^\circ - \angle BAC - \angle CBD$ Substituting the given values: $\angle ABC = 180^\circ - 50^\circ - 30^\circ = 100^\circ$

Step 2: Use the Angle Sum Property in Triangle $BCD$

Similarly, in triangle $BCD$, the sum of the angles is $180^\circ$. Therefore, $\angle BCD = 180^\circ - \angle CBD - \angle BDC$ Substituting the given values: $\angle BCD = 180^\circ - 30^\circ - 25^\circ = 125^\circ$

Step 3: Apply the Exterior Angle Theorem

The Exterior Angle Theorem states that an exterior angle of a triangle is equal to the sum of the two opposite interior angles. In this case, consider the exterior angle at $E$ in triangle $AEB$.

Since $\angle AEB$ is the exterior angle to $\triangle AEC$, it can be found by summing $\angle BAC$ and $\angle BDC$. Therefore: $\angle AEB = \angle BAC + \angle BDC = 50^\circ + 25^\circ = 75^\circ$

Conclusion:

The angle $\angle AEB$ is $75^\circ$.

Would you like further details or have any questions about the solution?

Here are 5 related questions you can explore:

1. How does the angle sum property help in solving quadrilateral problems?
2. What are the key differences between convex and concave quadrilaterals?
3. How does the Exterior Angle Theorem apply to other geometric shapes?
4. Can you find a relationship between diagonals and angles in cyclic quadrilaterals?
5. What role does the intersection point of diagonals play in determining angles in quadrilaterals?

Tip: Always check if a quadrilateral is cyclic when dealing with angle problems, as special angle properties apply in cyclic quadrilaterals.