Question

Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum
of the 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,

⇒ (AC) 2=(AE) 2+(CE) 2 [ By Pythagorasntheorem ]

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 )

Adding ( 1 ) and ( 2 ), we get

(AC) 2+(BD) 2=(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 )

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.

Hope it helped you.