Question

Prove that the sum of squares of sides of a parallelogram is equal to the sum of the squares of it's diagonals​

Answer 1

Question:-

• Prove that the sum of squares of sides of a parallelogram is equal to the sum of the squares of it's diagonals

Proof:-

In parallelogram ABCD, AB = CD, BC = AD

Construction:-

Draw perpendiculars from C and D on AB as shown.

Now:-

In right angled ΔAEC, AC² = AE² + CE² [By Pythagoras theorem]

⇒ AC² = (AB + BE)² + CE²

⇒ AC² = AB² + BE² + 2 AB × BE + CE²  ---(i)

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  ----(ii)

Add equation (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)  ------(iii)

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.

• Hence proved

Hope it helps you ✔️

Answer 2

$\color{grey}\underline{ \rm \large \: \: Statement :- }$

• The sum of squares of sides of a parallelogram is equal to the sum of the squares of it's diagonals.

$\color{grey}\underline{ \rm \large \: \: Given :- }$

• ABCD is a parallelogram .

$\color{grey}\underline{ \rm \large \: \: To \: prove :- }$

$\: \: \: \: \: \: \sf \star \: \: Sum \: of \: squares \: of \: diagonals \: = \: sum \: of \: squares \: of \: its \: sides$

$\: \: \: \: \: \: \sf \star \: \: AC² + BD² = AB² + BC² + CD² + DA²$

$\color{grey} \underline{ \rm \large \: \: Construction :- }$

• Draw AX is perpendicular to CD and BY is is extended to Y.

$\color{grey} \underline{ \rm \large \: \: Proof :- }$

In Right ∆ AXC ,

Applying Pythagoras theorem,

AC² = AX² + CX² ............................ ( 1 )

In Right ∆ BYD ,

Applying Pythagoras theorem,

BD² = BY² + DY² ............................ ( 2 )

Hence, The equations are

AC² = AX² + CX² ............................ ( 1 )

BD² = BY² + DY² ............................ ( 2 )

From equation 2 ,

BD² = BY² + DY²

Now, putting DY = CD + CY ,

$\\ \: \: \: \: \: \: \: \: \: \sf \: BD ^{2} = BY ^{2} + (CD + CY )^{2}$

$\\ \: \: \:\: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies\: \: \: \: \sf \: BD ^{2} = BY ^{2} + CY² + CD² + 2CD * CY$

$\\ \: \: \: \: \: \: \: \: \: \: \:\:\: \: : \implies \: \sf \: BD ^{2} =( BY ^{2} + CY ^{2}) + CD ^{2} + 2 CD \times CY$

$\\ \green{ \boxed{ \rm \: In \: ∆BYC , \: By \: Pythagoras \: theorem \: \: BC² = BY² + CY ²}}$

Therefore,

$\: \: \: \: \sf \underline{ \: \: BD² = BC² + CD² + 2CD×CY .......................(3)\: \: }$

From equation 1 ,

AC² = AX² + CX²

putting CX = CD - DX

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf { \longrightarrow} \: AC² = AX² + (CD - DX)²$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf { \longrightarrow} \: AC² = AX² + CD ^{2} - DX ² + 2CD \times DX$

$\\ \green{ \boxed{ \rm \: In \: ∆AXD , \: By \: Pythagoras \: theorem \: \: AD^{2} = AX^{2} + D X ^{2} } }$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf { \longrightarrow} \: AC² = AD² + CD ^{2} + 2CD \times DX$

$\\ \green{ \boxed{ \rm \: \: \: In \: parallelogram \: ABCD , \: opposite \: sides \: are \: equal , \: \: \therefore \: \: CD = AB } }$

$\\ \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf { \longrightarrow} \: \: \: \: \: AC² = AD² + AB ^{2} + 2CD \times DX \: ...................... \: (4)$

Hence, The equations are

$\\ \: \: \: \: \sf \: \: BD² = BC² + CD² + 2CD×CY .......................(3)\: \:$

$\\ \: \:\sf \: \: \: \: \: AC² = AD² + AB ^{2} + 2CD \times DX \: ...................... \: (4)$

Now,

Adding equation 3 & 4 ,

$\\ \: \: \: \: \: \: \: \sf \: BD² + AC² = BC² + 2CD.CY + AD² + AB² - 2CD.DX$

$\\ \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: \: \: \sf \: BD² + AC² = BC² + AD² + AB² + 2CD (CY - DX) ..................... \: (5)$

Now, we need to prove CY = DX ,

In ∆AXD and ∆BYC ,

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \angle \: AXD = \angle \: BYC \: ..............................(\angle \: AXD = \angle \: BYC \: = 90 \degree \: by \: construction)$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: AD = BC \: \: \: \: ........................ \: (opposite \: sides\: of \: parallelogram \: are \: equal)$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: AX = BY .......................... ( Perpendicular \: bisectors )$

$\\ \: \: \: \: \: \: \sf \: \pink{ ∆ AXD = ∆BYC \: ......................( SAS \: congruency )}$

Thus ,

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\blue{ \frak{ \: DX = CY \: \: ................... \: (CPCT)}}}$

Putting CY = DX in equation 5 ,

$\\ \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: \: \: \sf \: BD² + AC² = BC² + AD² + AB² + 2CD ( \bold{CY }- DX) ..................... \: (5)$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: \: \: \sf \: BD² + AC² = BC² + AD² + AB² + 2CD ( \bold{DX }\: - DX) ..................... \: (5)$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: \: \: \sf \: BD² + AC² = BC² + AD² + AB² + 2CD \times 0 ..................... \: (5)$

$\\ \: \: \: \: \: \: \: \: \: \color{navy}\underline{\boxed{\rm \: BD² + AC² = BC² + AD² + AB² ..................... \: (5)}}$

Hence proved ✅

Be brainly ♡