Proving the Identity: AB² + BC² + CD² + DA² = AC² + BD² in a Rhombus
In this article, we will prove the identity AB² + BC² + CD² + DA² = AC² + BD² for a rhombus ABCD.
Understanding the Problem:
A rhombus is a quadrilateral with all sides equal. Its diagonals bisect each other at right angles. The identity states that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Proof:
Let's break down the rhombus ABCD into four congruent right-angled triangles. Let O be the point of intersection of the diagonals AC and BD.
Key Properties:
- Sides: AB = BC = CD = DA
- Diagonals: AC and BD bisect each other at right angles. Therefore, AO = CO and BO = DO.
- Right Angles: ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°.
Applying the Pythagorean Theorem:
Using the Pythagorean Theorem on each of the four right-angled triangles, we get:
- AB² = AO² + BO²
- BC² = BO² + CO²
- CD² = CO² + DO²
- DA² = DO² + AO²
Adding all these equations together, we get:
AB² + BC² + CD² + DA² = 2AO² + 2BO² + 2CO² + 2DO²
Now, since AO = CO and BO = DO, we can simplify the equation to:
AB² + BC² + CD² + DA² = 2(AO² + BO²) + 2(CO² + DO²)
This simplifies further to:
AB² + BC² + CD² + DA² = 2(AC²/4) + 2(BD²/4)
Final Step:
Simplifying the equation, we get:
AB² + BC² + CD² + DA² = AC² + BD²
Conclusion:
We have successfully proved that in a rhombus ABCD, the sum of the squares of the sides (AB² + BC² + CD² + DA²) is equal to the sum of the squares of its diagonals (AC² + BD²).
