Question

Prove that in a parallelogram sum of the squares of the sides is equal to sum of the squares of the diagonals.

Answer 1

Hi dear friend..

See the attached file.. There is figure..

SOLUTION is here

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.

I HÖPÉ IT WILL HELP Ýøü

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Answer 2

Answer:

Step-by-step explanation:

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.