Question

in the rhombus ABCD show that 4AB^2 =AC^2+ BD^2

Answer 1

Hey there!

Here, We see that,
AO = 1/2AC
BO = 1/2BD

In triangle ABO,

According to Pythagoras theorem,

AO² + BO² = AB²

(1/2AC)² + (1/2BD)² = AB²

1/4AC² + 1/4BD² = AB²

1/4 ( AC² + BD²) = AB²

(AC² + BD²) = 4AB²

[ Hence proved ]