Question

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

Plzzzzzz

Answer 1

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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Answer 2

Answer:

QED

Step-by-step explanation:

ABCD is parrallelogram

AD || BC & AD= BC

AB || CD & AB = CD

Let draw perpendicular DE from D at AB  ( Assume E lies on AB)

Let draw perpendicular CF from C at AB  ( then F lies on extension of AB)

DE = CF  , AE = BF as triangle AED similar to triangle BFC where AD = BC (as perpendicular drawn from a parallelogram)

AC² = (AB + BF)² + CF²

AC² = AB² + BF² + 2AB.BF + CF²

where BC² = BF² + CF²

AC² = AB² + BC² + 2AB.BF   - Eq 1

BD² = (AB - AE)² + DE²

BD² = AB² + AE² - 2AB.AE + DE²

where AD² = AE² + DE²

BD² = AB² + AD² - 2AB.AE  - Eq2

Now adding eq1 & Eq2

AC² + BD² = AB² + BC² + 2AB.BF + AB² + AD² - 2AB.AE

AC² + BD² =  AB² + BC² + AB² + AD² + 2AB.BF - 2AB. AE

BF = AE  & AB = CD

AC² + BD² =  AB² + BC² + CD² + AD² + 2AB.AE - 2AB. AE

AC² + BD² =  AB² + BC² + CD² + AD²

So sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

QED