Question

Prove that the sum of squares of sides of a parallelogram is equal to the sum of the squares of it's diagonals.​

Answer 1

Draw perpendiculars from C and D on AB as shown. Thus the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

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Answer 2

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 )

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 )

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.