Prove that sum of squares of diagonals of a parellogram is equal to the sum of squares of its sides.

Prove that sum of squares of diagonals of a parellogram is equal to th

1.	Prove that sum of squares of diagonals of a parellogram is equal to the sum of squares of its sides.
Answer» Given: ABCD is a parallelogram whose diagonals are AC and BD.To prove: AB2 + BC2 + CD2 + DA2 = AC2 + BD2Construction: Draw AM {tex} \\bot {/tex} DC and BN{tex} \\bot {/tex} D(Produced)Proof: In right triangle AMD and BNC.AD = BC ..............Opp.sides of a \|\|gmAM = BN ............Both are altitudes of the same parallelogram to the same base{tex}\\therefore {/tex}\u200b{tex}\\triangle {/tex}\u200b AMD {tex}\\cong{/tex}\u200b{tex}\\triangle {/tex}\u200bBNC ................RHS congruence criterion{tex}\\therefore {/tex} MD = NC .........(1).........CPCTIn right triangle BND,{tex}\\because {/tex}{tex}\\angle{/tex} N=90°{tex}\\therefore {/tex} BD2 = BN2 + DN2 .............By Pythagoras theorem= BN2 + (DC + CN)2= BN2 + DC2 + CN2 + 2DC.CN= (BN2 + CN2) + CN2 + 2DC.CN= BC2 + DC2 + 2DC.CN ..........(2)In right triangle BNC with {tex}\\angle{/tex}N = 90oBN2+CN2 = BC2 ......By Pythagoras theoremIn right triangle AMC{tex}\\because {/tex}{tex}\\angle{/tex} M=90o{tex}\\therefore {/tex} AC2 = AM2 + MC2= AM2 = (DC - DM)2= AM2 + DC2 + DM2 - 2DC.DM= (AM2 + DC2) + DC2 - 2DC.DM= AD2 + DC2 - 2DC.DM{tex}\\because {/tex} In right triangle AMD with {tex}\\angle{/tex} M=90°AD2 = AM2 + DM2 ..........[By Pythagoras theorem]= AD2 + AB2 - DC.CN .......From(1)Adding (3) and (2) ,we getAC2 + BD2 = (AD2 + AB) + (BC2 + DC) = AB2 + BC2 + BC2 + CD2 + DA2