if ABCD is a rhombus then prove that 4 a b square is equals to AC square + BD square
Answers
Given : ABCD is a rhombus
To Find : prove that 4 AB² = AC² + BD²
Solution:
Property of rhombus
Diagonal of rhombus bisects each other perpendicularly
and all sides of rhombus are equal
Let say Say side of Rhombus = a
AB = BC = CD = AD = a
and diagonal are d₁ & d₂
d₁ = AC
d₂ = BD
d₁ & d₂ bisect perpendicularly at O
Applying Pythagoras theorem
(d₁/2)² + (d₂/2)² = a²
=> d₁²/4 + d₂²/4 = a²
Multiplying by 4 both sides
=> d₁² + d₂² = 4a²
=> AC² + BD² = 4AB²
QED
Hence Proved
4 times the square of any side = sum of squares of its diagonals
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