In a rhombus ABCD, prove that AC² + BD² = 4AB².

Question

KnowledgeBoat · Accepted Answer

In rhombus, AB = BC = CD = AD (All sides of a rhombus are equal) Diagonal AC and BD intersect at point O. Diagonals bisect each other at right angles. So, AO = OC = × AC and BO = OD = × BD In right angled △AOB, Using Pythagoras theorem, Hypotenuse2 = Perpendicular2 + Base2 ⇒ AB2 = AO2 + BO2 ∴ 4AB2 = AC2 + BD2 Hence, proved that AC2 + BD2 = 4AB2.

Mathematics

In a rhombus ABCD, prove that AC² + BD² = 4AB².

Pythagoras Theorem

Related Questions

In △ABC, ∠B = 90° and D is the mid-point of BC. Prove that
(i) AC² = AD² + 3 CD²
(ii) BC² = 4 (AD² - AB²)

Two poles of height 9 m and 14 m stand vertically on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.

Mathematics

In a rhombus ABCD, prove that AC2 + BD2 = 4AB2.

Pythagoras Theorem

Related Questions

In △ABC, ∠B = 90° and D is the mid-point of BC. Prove that(i) AC2 = AD2 + 3 CD2(ii) BC2 = 4 (AD2 - AB2)

Two poles of height 9 m and 14 m stand vertically on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.

In a rhombus ABCD, prove that AC² + BD² = 4AB².

In △ABC, ∠B = 90° and D is the mid-point of BC. Prove that
(i) AC² = AD² + 3 CD²
(ii) BC² = 4 (AD² - AB²)