Question

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

Answer 1

In parallelogram $ABCD$,

$AB=CD$ and $BC=AD$

Draw perpendiculars from $C$ and $D$ on $AB$ as shown.

In right angled $△AEC,$

$\Rightarrow$ $(AC{)}^2=(AE{)}^2+(CE{)}^2$ [ By Pythagoras theorem ]

$\Rightarrow$ $(AC{)}^2=(AB+BE{)}^2+(CE{)}^2$

$\Rightarrow$ $(AC{)}^2=(AB{)}^2+(BE{)}^2+2\times AB\times BE+(CE{)}^2$ ----- ( 1 )

From the figure $CD=EF$ [ Since $CDFE$ is a rectangle ]

But $CD=AB$

$\Rightarrow$ $AB=CD=EF$

Also $CE=DF$ [ Distance between two parallel lines

$\Rightarrow$ $△AFD\cong △BEC$ [ RHS congruence rule ]

$\Rightarrow$ $AF=BE$ [ CPCT ]

Considering right angled $△DFB$

$\Rightarrow$ $(BD{)}^2=(BF{)}^2+(DF{)}^2$ [ By Pythagoras theorem ]

$\Rightarrow$ $(BD{)}^2=(EF-BE{)}^2+(CE{)}^2$ [ Since $DF=CE$ ]

$\Rightarrow$ $(BD{)}^2=(AB-BE{)}^2+(CE{)}^2$ [ Since $EF=AB$ ]

$\Rightarrow$ $(BD{)}^2=(AB{)}^2+(BE{)}^2-2\times AB\times BE+(CE{)}^2$ ----- ( 2 )

Adding ( 1 ) and ( 2 ), we get

$(AC{)}^2+(BD{)}^2=(AB{)}^2+(BE{)}^2+2\times AB\times BE+(CE{)}^2+(AB{)}^2+(BE{)}^2-2\times AB\times BE+(CE{)}^2$

$\Rightarrow$ $(AC{)}^2+(BD{)}^2=2(AB{)}^2+2(BE{)}^2+2(CE{)}^2$

$\Rightarrow$ $(AC{)}^2+(BD{)}^2=2(AB{)}^2+2[(BE{)}^2+(CE{)}^2]$ ---- ( 3 )

In right angled $△BEC,$

$\Rightarrow$ $(BC{)}^2=(BE{)}^2+(CE{)}^2$ [ By Pythagoras theorem ]

Hence equation ( 3 ) becomes,

$\Rightarrow$ $(AC{)}^2+(BD{)}^2=2(AB{)}^2+2(BC{)}^2$

$\Rightarrow$ $(AC{)}^2+(BD{)}^2=(AB{)}^2+(AB{)}^2+(BC{)}^2+(BC{)}^2$

$\therefore$ $(AC{)}^2+(BD{)}^2=(AB{)}^2+(BC{)}^2+(CD{)}^2+(AD{)}^2$

$\therefore$ The sum of the squares of the diagonals of a parallelogram is equal to the sum of squares of its sides.