Question

In rhombus ABCD, prove that 4AB square=AC square+BD square
PLZ give an answer with detailed explanation! !

Answer 1

let o is a point where the diagonals cut
consider triangle AOB
AB^2=AO^2+OB^2 ( from Pythagoras theory)
consider triangle AOD
AD^2=AO ^2+OD^2(")
consider triangle DOC
DC^2=OD^2+OC^2(")
consider triangle BOC
BC^2=OB^2+OC^2(")

and we know that AB=BC=CD=AD. And AO+OC=AC,OB+OD=BD,AO=OC,OB=OD
add all the equations
AB^2+BC^2+CD^2+AD^2=2(AO^2 )+2(OB^2)+2(OC^2)+2(OD^2)
(4AB)^2=2(AO+OC)^2+2(OB+OD)^2
16AB^2=(2AC^2)+(2BD^2)
16AB^2=2^2{AC^2+BD^2}
16AB^2=4(AC^2+BD^2)
4AB^2=AC^2+BD^2
hence proved


where x^2=x squared

Answer 2

hope it helps all.......$$$