Question

plsss answer this question❤❤❤❤❤❤❤​

Answer 1

Answer:⤵️

given that ABCD is A PARALELLOGRAM

we know that diagrams of a paralellogram intersect each other

$Ao = \frac{1}{2}Ac \: and \: Bo = \frac{1}{2}BD$

so in ABO , angleABO=90⁰

so by phythagoras thm in∆ABO

AO²=BO²+AB²

=> AB²=AO²-BO²

$(\frac{1}{2}AC^2) - ( \frac{1}{2}BD^2)$

$\frac{1}{4}AC^2 - \frac{1}{4}BD^2$

$\frac{1}{4} (AC^2-BD^2)$

Ac²-BD²=4AB²