Prove that the sum of squares of sides of a parallelogram is equal to the sum of the squares of it's diagonals ?
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Answers
Answer: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.
Step-by-step explanation: