Question

Prove that in any parallelogram, the sum of the squares of all sides is equal to the sum of the squares of the diagonals.

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Answer 1

Answer:

i.e. O is the mid point of AC and BD. In ∆ABD, point O is the midpoint of side BD. In ∆CBD, point O is the midpoint of side BD. Hence, the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

...

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.

Answer 2

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Prove that in any parallelogram, the sum of the squares of all sides is equal to the sum of the squares of the diagonals.

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In parallelogram ABCD,

AB = CD, BC = AD Draw perpendiculars from C and D on AB as shown.

In right angled ΔAEC,

AC² = AE² + CE²

[By Pythagoras theorem]

⇒ AC² = (AB + BE)² + CE²

⇒ AC²= AB² + BE2 + 2AB × BE + CE² → (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)

ΔAFD ≅ ΔBEC (RHS congruence rule)

⇒ AF = BE

Consider right angled ΔDFB

BD² = BF² + DF² [By Pythagoras theorem]

= (EF – BE)² + CE² [Since DF = CE]

= (AB – BE)2 + CE² [Since EF = AB]

⇒ BD² = AB² + BE² – 2AB × BE + CE² → (2)

Add (1) and (2),

we get AC² + BD² = (AB² + BE² + 2AB × BE + CE²)+ (AB² + BE² – 2AB × BE + CE²)

= 2AB² + 2BE² + 2CE² AC² + BD²

= 2AB² + 2(BE² + CE²) → (3)

From right angled ΔBEC,

BC² = BE² + CE² [By Pythagoras theorem]

Hence equation (3) becomes,

AC² + BD² = 2AB² + 2BC²

= AB² + AB² + BC2 + BC²

= AB² + CD² + BC² + AD²

∴ AC² + BD² = AB² + BC² + CD² + AD²

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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