ABCD is a quadrilateral in which AC and BD are its diagonals. Which of the following inequalities is always true for any quadrilateral?

Introduction / Context:
This question explores the relationship between the sides and diagonals of a quadrilateral. It asks which inequality always holds, regardless of the specific shape of the quadrilateral. This is an extension of the triangle inequality to quadrilaterals and tests your understanding of how distances behave in polygons. Such conceptual questions are often used in aptitude and geometry exams to probe reasoning rather than computation.

Given Data / Assumptions:

- ABCD is any quadrilateral, not necessarily convex or cyclic, but usually considered simple and non self intersecting.
- AC and BD are its diagonals, joining opposite vertices.
- AB, BC, CD and DA are the four side lengths of the quadrilateral.
- We want to know how the sum of the side lengths compares with the sum of the diagonal lengths AC and BD.
- No special angle or side length conditions are given, so the statement must hold for all quadrilaterals.

Concept / Approach:
The triangle inequality states that in any triangle, the sum of the lengths of any two sides is greater than the length of the third side. We can apply this idea to triangles formed inside the quadrilateral by its diagonals. For instance, diagonal AC splits the quadrilateral into triangles ABC and ADC. Applying triangle inequalities in these triangles yields inequalities involving AB, BC, AC and also AD, CD, AC. Adding these inequalities allows comparison of the perimeter with 2AC, and similar reasoning with diagonal BD shows that the total perimeter is greater than the sum of diagonals.

Step-by-Step Solution:
Step 1: Consider diagonal AC. It divides quadrilateral ABCD into triangles ABC and ADC.Step 2: In triangle ABC, triangle inequality gives AB + BC > AC.Step 3: In triangle ADC, triangle inequality gives AD + DC > AC.Step 4: Add the two inequalities: (AB + BC) + (AD + DC) > AC + AC = 2AC.Step 5: Thus AB + BC + CD + DA > 2AC.Step 6: Similar reasoning using diagonal BD and triangles ABD and CBD yields AB + BC + CD + DA > 2BD.Step 7: From the two results, it follows that AB + BC + CD + DA is greater than both 2AC and 2BD, and therefore greater than AC + BD as well. So AB + BC + CD + DA > AC + BD is always true.

Verification / Alternative check:
Consider a simple example like a square of side 1 unit. Its perimeter is 4 units. The diagonal length is square root of 2 units, so AC + BD = 2 * square root of 2 which is approximately 2.828. Clearly 4 is greater than 2.828. You can also test with a rectangle of sides 2 and 1: perimeter is 6, diagonal is square root of 5, so AC + BD = 2 * square root of 5 which is about 4.472. Again 6 is greater than 4.472. These examples support the general inequality.

Why Other Options Are Wrong:
The inequality AB + BC + CD + DA < AC + BD contradicts even simple examples such as a square. Equality AB + BC + CD + DA = AC + BD or AB + BC + CD + DA = 2(AC + BD) would require special rigid conditions that do not hold for arbitrary quadrilaterals. The stronger inequality AB + BC + CD + DA > 2(AC + BD) is also not guaranteed; for some shapes the sum of diagonals can be relatively large compared to the sides, so this statement need not always be true.

Common Pitfalls:
A frequent mistake is to assume that diagonals might be longer when connecting distant vertices, leading to the belief that the sum of diagonals could exceed the perimeter. Another error is to misapply triangle inequality or to think it only holds for identical triangles. Remember that each diagonal creates two triangles, and applying the triangle inequality in each gives solid bounds relating sides and diagonals.

Final Answer:
The inequality that always holds is AB + BC + CD + DA > AC + BD.