Given: ABCD is a quadrilateral and AC is one of its diagonals.

To Prove: AB + BC + CD > DA

Proof:

Step 1: Understand the properties of a quadrilateral.

A quadrilateral is a polygon with four sides. The sum of the lengths of any two sides of a quadrilateral is always greater than the length of the third side. This property is known as the Triangle Inequality Theorem.

Step 2: Analyze the given quadrilateral ABCD.

Let's consider the quadrilateral ABCD, where AC is one of its diagonals. To prove AB + BC + CD > DA, we need to show that the sum of the lengths of AB, BC, and CD is greater than the length of DA.

Step 3: Divide into two triangles.

We can divide the quadrilateral ABCD into two triangles, namely triangle ABC and triangle CDA, by drawing the diagonal AC.

Step 4: Apply the Triangle Inequality Theorem to each triangle.

In triangle ABC, we have:
AB + BC > AC (Triangle Inequality Theorem)

In triangle CDA, we have:
CD + DA > AC (Triangle Inequality Theorem)

Step 5: Add both inequalities.

By adding the two inequalities, we get:
AB + BC + CD + DA > AC + AC

Step 6: Simplify the equation.

Since AC + AC = 2AC, we can simplify the inequality as:
AB + BC + CD + DA > 2AC

Step 7: Substitute the value of AC.

Since AC is the diagonal of the quadrilateral ABCD, it can be expressed in terms of the sides of the quadrilateral. Let's say AC = p + q, where p and q are the lengths of the two opposite sides of the quadrilateral.

Substituting the value of AC, we have:
AB + BC + CD + DA > 2(p + q)

Step 8: Simplify further.

By simplifying the inequality, we get:
AB + BC + CD + DA > 2p + 2q

Step 9: Rearrange the terms.

Rearranging the terms, we have:
AB + BC + CD + DA > (AB + DA) + (BC + CD)

Step 10: Conclusion.

From the rearranged inequality, we can observe that (AB + DA) represents the sum of the lengths of the opposite sides of the quadrilateral, and (BC + CD) represents the sum of the lengths of the other two opposite sides.

Therefore, AB + BC + CD + DA is greater than the sum of the lengths of the opposite sides of the quadrilateral, which proves that AB + BC + CD > DA.

Hence, the statement is proved.