Given the image, we have a square $ABCD$ with a triangle $AEF$ inside it. Here are the key details we can observe:

1. $ABCD$ is a square, so each internal angle is $90^\circ$.
2. In triangle $AEF$:
   • $\angle EAF = 45^\circ$
   • $\angle AFE = 70^\circ$
   • We need to find $\angle AEF$, which we'll denote as $x$.

Using the property that the sum of the angles in a triangle is $180^\circ$: $\angle EAF + \angle AFE + \angle AEF = 180^\circ$ Substitute the known angles: $45^\circ + 70^\circ + x = 180^\circ$ $115^\circ + x = 180^\circ$ Subtract $115^\circ$ from both sides to find $x$: $x = 180^\circ - 115^\circ$ $x = 65^\circ$

So, $\angle AEF = 65^\circ$.

Would you like more details on this solution or have any questions?

Here are some related questions you might find interesting:

1. How would the solution change if $ABCD$ was not a square but a rectangle?
2. What if one of the angles in triangle $AEF$ was different? How would that affect $x$?
3. How can we use the properties of a square to find unknown angles in other geometric shapes?
4. What are some other ways to solve for unknown angles in polygons?
5. How does knowing the properties of triangles help in solving problems involving quadrilaterals?

Tip: Always double-check the sum of angles in polygons. For a triangle, they should always sum to $180^\circ$, which is a useful property for solving many geometry problems.