Question

HII GUYZZ....

HERE'S THE QUESTION OF THE DAY!

QUE:-

PROVE THAT THE SUM OF THE SQUARE OF THE DIAGOÑAL OF A PARALLELOGRAM IS EQUAL TO THE SUM OF THE SQUARE OF ITS SIDES.

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❌❌SPAMMERS BE AWAY❌❌

Answer 1

⇒ 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

_______Heyy Buddy __________

_____Here's your Answer _______

Given :-

ABCD is a Parallelogram with Diagonal AC & BD.

To Prove :-

${ab}^{2} + {bc}^{2} + {cd}^{2} + {ad}^{2} = {ac}^{2} + {bd}^{2}$

Proof :-



We know that if AD is a median of ABC , then

${ab}^{2} + {ac}^{2} = 2 {ad}^{2} + \frac{1}{2} {bc}^{2}$

Since,

Diagonal of Parallelogram bisect each other.

Therefore, BO & DO r median of Triangle ABC & ADC respectively.

Now,

${ab}^{2} + {bc}^{2} = 2 {ob}^{2} + \frac{1}{2} {ac}^{2} ........(1) \\ \\ and \\ \\ {ad}^{2} + {cd}^{2} = 2 {od}^{2} + \frac{1}{2} {ac}^{2} .........(2)$


Adding equation 1 and 2. we get,

${ab}^{2} + {bc}^{2} + {cd}^{2} + {ad}^{2} = 2 ( {ob}^{2} + {od}^{2} ) + {ac}^{2} \\ \\ = > {ab}^{2} + {bc}^{2} + {cd}^{2} + {ad}^{2} = 2( \frac{1}{4} {bd}^{2} + \frac{1}{4} {bd}^{2} ) + {ac}^{2} \\ \\ = > {ab}^{2} + {bc}^{2} + {cd}^{2} + {ad}^{2} = {ac}^{2} + {bd}^{2}$

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