Q24 of 95 Page 5

In a rhombus ABCD, prove that: AB2 + BC2 + CD2 + DA2 = AC2 + BD2


In rhombus ABCD, AB = BC = CD = DA


We know that diagonals bisect each other at 90°


And


Consider right triangle AOB


(Perpendicular)2 + (Base)2 = (Hypotenuse)2


⇒ (OA)2 + (OB)2 = (AB)2



⇒ AC2 + BD2 = 4AB2


⇒ AC2 + BD2 = AB2 + AB2 + AB2 + AB2


⇒ AC2 + BD2 = AB2 + BC2 + CD2 + DA2


Hence Proved


More from this chapter

All 95 →
22

In an isosceles ΔABC, AB = AC and BD ⊥ AC. Prove that BD2 — CD2 = 2CD AD.

 23

In a quadrilateral, ΔBCD, ∠B = 90°. If AD2= AB2 + BC2 + CD2, prove that ∠ACD = 90°.

 25

In an equilateral triangle ABC, AD is the altitude drawn from A on side BC. Prove that 3AB2= 4AD2.

 26

In ΔABC, AB = AC. Side BC is produced to D. Prove that (AD2 —AC2) = BD . CD