Question

33 Prove that the sum of the squares of the diagonals of a parallelogram is equal to the
sum of the squares of its sides. ​

Answer 1

$\bf\large\underline\pink{Answer:-}$

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.

Answer 2

Answer:

sorry I don't know. this problem because I am 6 th