Math, asked by aqdasrahman4739, 10 months ago

Prove That the sum of the diagonals of the parallelogram is equal to the sum of the squares of its sides

Answers

Answered by KaushalKishorKumar21
0

Answer:

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.

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