Question

Prove that in any parallelogram, the sum of the squares of all sides is equal to the sum of the squares of the diagonals. DON'T SPAM IT'S URGENT. . . ​

Answer 1

i.e. O is the mid point of AC and BD. In ∆ABD, point O is the midpoint of side BD. In ∆CBD, point O is the midpoint of side BD. Hence, 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

$\huge\sf{\underline{\underline{\pink{Solution}}}}$

In parallelogram ABCD, AB = CD, BC = AD Draw perpendiculars from C and D on AB as shown.

$\sf{\bold{In \: right \: angled \: Δ,}}$

$\sf{\bold{AEC, {AC}^{2} = {AE}^{2} + {CE}^{2} [By Pythagoras theorem]}}$

$\implies\sf{\bold{ {AC}^{2} = { (AB + BE)}^{2} + {CE}^{2} }}$

$\implies\sf{\bold{ { AC}^{2} = {AB}^{2} + {BE}^{2} + 2 AB × BE + {CE}^{2} }}$

(1) From the figure CD = EF (Since CDFE is a rectangle)

$\sf{\bold{But \: CD= AB => AB = CD = EF}}$

$\sf{\bold{Also \: CE = DF}}$

(Distance between two parallel lines)

ΔAFD ≅ ΔBEC (RHS congruence rule) => AF = BE.

Consider right angled ΔDFB

$\sf{\bold { {BD}^{2} = {BF}^{2} + {DF}^{2} [By Pythagoras theorem]}}$

$\sf{\bold {= {(EF – BE)}^{2} + {CE}^{2} [Since DF = CE] }}$

$\sf{\bold {= {(AB – BE)}^{2} + {CE}^{2} [Since EF = AB] }}$

$\implies\sf{\bold{ {BD}^{2} = {AB}^{2} + BE2 – 2 AB × BE + {CE}^{2} → (2)}}$

Add (1) and (2), we get,

$\sf{\bold{ {AC}^{2} + {BD}^{2} </p><p>= ( {AB}^{2} + {BE}^{2} + 2AB × BE + {CE}^{2} ) + ( {AB}^{2} + {BE}^{2} – 2 AB × BE + {CE}^{2} )}}$

$\sf{\bold{= {2AB}^{2} + {2BE}^{2} + {2CE}^{2} }}$

$\sf{\bold{ {AC}^{2} + {BD}^{2} = {2AB}^{2} + 2( {2BE}^{2} + {CE}^{2} )}} \: \:-> (3)$

From right angled ΔBEC, BC2 = BE2 + CE2 [By Pythagoras theorem].

Hence equation (3) becomes,

$\sf{\bold{ {AC}^{2} + {BD}^{2} = {2AB}^{2} + { 2BC}^{2} } }$

$\sf{\bold{= {AB}^{2} + {AB}^{2} + {BC}^{2} + {BC}^{2} }}$

$\sf{\bold{= {AB}^{2} + {CD}^{2} + {BC}^{2} + {AD}^{2} }}$

$\sf{\bold{∴ {AC}^{2} + {BD}^{2} = {AB}^{2} + {BC}^{2} + {CD}^{2} + {AD}^{2} }}$

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