Math, asked by manasaraniganta64, 3 months ago

sum of the squares of the diogonals of parallelogram is equal to the sum of the squares of its sides​

Answers

Answered by ItzBrainlyGirl024
4

Answer:

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 + 2 AB × 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 – 2 AB × BE + CE2  → (2)

Add (1) and (2), we get

AC2 + BD2 = (AB2 + BE2 + 2 AB × BE + CE2) + (AB2 + BE2 – 2 AB × 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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Answered by tabrikulislam
1

Let □ABCD be a parallelogram.

Let its diagonals AC and BD intersect at O.

In △ABC,

BO is the median ....Diagonals of a parallelogram bisect each other

∴ By Apollonius theorem,

AB

2

+BC

2

=2OB

2

+2OA

2

....(1)

In △ADC,

DO is the median ....Since diagonals bisect each other

∴ By Apollonius theorem,

AD

2

+DC

2

=2OD

2

+2OC

2

....(2)

Adding (1) and (2) we get,

AB

2

+BC

2

+AD

2

+DC

2

=2OB

2

+2OA

2

+2OD

2

+2OC

2

AB

2

+BC

2

+CD

2

+AD

2

=2OB

2

+2OA

2

+2OB

2

+2OA

2

AB

2

+BC

2

+CD

2

+AD

2

=4OB

2

+4OA

2

AB

2

+BC

2

+CD

2

+AD

2

=4(

2

1

×DB)

2

+4(

2

1

×CA)

2

[∵OB=

2

DB

andOA=

2

CA

]

AB

2

+BC

2

+CD

2

+AD

2

=4(

4

1

×DB

2

)+4(

4

1

×CA

2

)

AB

2

+BC

2

+CD

2

+AD

2

=DB

2

+CA

2

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