Question

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

Answer 1

Answer:

AC^2 + BD^2 = 4AB^2

Step-by-step explanation:

We know that the diagonals of rhombus bisect each other into two equal parts at 90°.

Therefore, by Pythagoras theorem,

= > ( half of 1st diagonal )^2 + ( half of 2nd  diagonal )^2 = side^2 or AB^2

In the given question,

= > ( AC / 2 )^2 + ( BD /2 )^2 = AB^2

= > AC^2 / 4 + BD^2 / 4 = AB^2

= > ( AC^2 + BD^2 ) / 4 = AB^2

= > AC^2 + BD^2 = 4 x AB^2

= > AC^2 + BD^2 = 4AB^2

Hence proved.