Problem Statement:
ABCD is a square. E, F, G, and H are the midpoints of sides AB, BC, CD, and DA respectively, such that AE = BF = CG = DH. We need to prove that EFGH is also a square.

Proof:

Step 1: Establishing the properties of a square

To prove that EFGH is a square, we need to establish the following properties of a square:

1. All sides are equal in length.
2. All angles are right angles.
3. Diagonals are equal in length and bisect each other at right angles.

Step 2: Proving equal side lengths

Let's consider the given information that AE = BF = CG = DH.

Since E and F are midpoints of AB and BC respectively, we can apply the midpoint theorem, which states that the line segment joining two midpoints of a triangle is parallel to the third side of the triangle and half of its length.

Therefore, EF || AD and EF = (1/2)AD.

Similarly, GH || AD and GH = (1/2)AD.

Now, we can see that AE = EF + FA.
Using the above information, AE = (1/2)AD + (1/2)AD = AD.

Similarly, BF = BC = AD.

Thus, all sides of EFGH are equal in length.

Step 3: Proving right angles

To prove that all angles of EFGH are right angles, we will use the property that the diagonals of a square are perpendicular bisectors of each other.

Let's consider the diagonals of ABCD: AC and BD.

Since E and G are midpoints of AC and CD respectively, EG is the perpendicular bisector of AC. Similarly, FH is the perpendicular bisector of BD.

We know that perpendicular bisectors of a line segment intersect at right angles.

Therefore, EG ⊥ AC and FH ⊥ BD.

Now, since AC and BD are diagonals of ABCD, and EG and FH are perpendicular bisectors of these diagonals, we can conclude that EG and FH are diagonals of EFGH.

Thus, all angles of EFGH are right angles.

Step 4: Proving equal diagonal lengths

To prove that the diagonals of EFGH are equal in length, we will use the property that the diagonals of a parallelogram bisect each other.

We know that EG is a diagonal of EFGH, and it bisects AC at point M (as E and G are midpoints of AC and CD respectively).

Similarly, FH is a diagonal of EFGH, and it bisects BD at point N (as F and H are midpoints of AB and DA respectively).

Since EG and FH bisect AC and BD respectively, we can conclude that EG and FH bisect each other at point P (as M and N are the midpoints of AC and BD respectively).

Therefore, EP = PG and FP = PH.

Thus, the diagonals of EFGH are equal in length.

Step 5: Conclusion

Based on the above proofs, we have established that EFGH satisfies all the properties of a square:

1. All sides are equal in length.
2. All angles are