Question

prove that the sum of squares of daigonals of a parallelogram is equal to the sum of the squares of its sides​

Answer 1

$\huge\star\:\:{\orange{\underline{\purple{\mathcal{Proof}}}}}$

In parallelogram ABCD,

AB=CD and BC=AD

Draw perpendiculars from C and D on AB as shown.

In right angled △AEC,

⇒ (AC) 2 =(AE) 2 +(CE) 2

[ By Pythagoras theorem ]

⇒ (AC) 2 =(AB+BE) 2 +(CE) 2

⇒ (AC) 2 =(AB) 2 +(BE) 2 +2×AB×BE+(CE) 2

----- ( 1 )

From the figure CD=EF

[ Since CDFE is a rectangle ]

But CD=AB

⇒ AB=CD=EF

Also CE=DF

[ Distance between two parallel lines]

⇒ △AFD≅△BEC [ RHS congruence rule ]

⇒ AF=BE [ CPCT ]

Considering right angled △DFB

⇒ (BD) 2 =(BF) 2 +(DF) 2

[ By Pythagoras theorem ]

⇒ (BD) 2

=(EF−BE) 2 +(CE) 2

[ Since DF=CE ]

⇒ (BD) 2 =(AB−BE) 2 +(CE) 2

[ Since EF=AB ]

⇒ (BD) 2 =(AB) 2 +(BE) 2 −2×AB×BE+(CE) 2

----- ( 2 )

${\red{\boxed{\large{\bold{Adding ( 1 ) and ( 2 ) we get}}}}}$

(AC) 2 +(BD) =(AB) 2 +(BE) 2 +2×AB×BE+(CE) 2(AB)2 +(BE) 2 −2×AB×BE+(CE) 2

⇒ (AC) 2 +(BD) 2 =2(AB) 2 +2(BE) 2 +2(CE) 2

⇒ (AC) 2 +(BD) 2 =2(AB) 2 +2[(BE) 2 +(CE) 2 ] ---- ( 3 )

In right angled △BEC,

⇒ (BC) 2 =(BE) 2 +(CE) 2

[ By Pythagoras theorem ]

Hence equation ( 3 ) becomes,

⇒ (AC) 2 +(BD) 2 =2(AB) 2 +2(BC) 2

⇒ (AC) 2 +(BD) 2 =(AB) 2 +(AB) 2 +(BC) 2 +(BC) 2

∴ (AC) 2 +(BD) 2 =(AB) 2 +(BC) 2 +(CD) 2 +(AD) 2

∴ The sum of the squares of the diagonals of a parallelogram is equal to the sum of squares of its sides.