Question

if ABCD IS A RHOMBUS,SHOW THAT AC SQUARE +BD SQUARE=4AB SQUARE
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Answer 1

I hope it will help you

Answer 2

Answer:

Step-by-step explanation:

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

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