Prove that the sum of the …

Given: ABCD is a parallelogram whose diagonals are AC and BD.
To prove: AB² + BC² + CD² + DA² = AC² + BD²
Construction: 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 ||gm
AM = BN ............Both are altitudes of the same parallelogram to the same base
{tex}\therefore {/tex} ​{tex}\triangle {/tex}​ AMD {tex}\cong{/tex} ​{tex}\triangle {/tex}​ BNC ................RHS congruence criterion
{tex}\therefore {/tex} MD = NC .........(1).........CPCT
In 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 = 90^o
BN²+CN² = BC² ......By Pythagoras theorem
In right triangle AMC
{tex}\because {/tex} {tex}\angle{/tex} M=90^o
{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 get
AC² + BD² = (AD² + AB) + (BC² + DC) = AB² + BC² + BC² + CD² + DA²