Question

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

Answer 1

Solution:-

$\sf\ In\: parallelogram\: ABCD,$

$\sf\ AB=CD\: and\: BC=AD$

$\sf\ Draw\: perpendiculars\: from\: C\: and\: D\: on$

$\sf\ AB\: as\: shown.$

In right angled △AEC,

⇒$\sf\ (AC)^2 = (AE)^2 + (CE)^2$

$\sf\ [ By\: Pythagoras\: theorem ]$

⇒ $\sf\ (AC)^2 = (AB+BE)^2 + (CE)^2$

⇒ $\sf\ (AC)^2 = (AB)^2 + (BE)^2 + 2 × AB × BE$

$\sf\ + (CE)^2-- ( 1 )$

From the figure CD = EF

$\sf\ [Since\: CDFE\: is \:a\: rectangle]$

$\sf\ But\: CD=AB$

⇒ $\sf\ AB=CD=EF$

$\sf\ Also \:CE=DF$

$\sf\ [Distance\: between\: two\: parallel\: lines]$

⇒ $\sf\ △AFD≅△BEC [RHS \:congruence\: rule ]$

⇒ $\sf\ AF=BE [ CPCT ]$

Considering right angled △DFB

⇒ $\sf\ (BD)^2 = (BF)^2 + (DF)^2$

$\sf\ [ By\: Pythagoras\: theorem]$

⇒$\sf\ (BD)^2 = (EF−BE)^2 + (CE)^2$

$\sf\ [ Since DF=CE ]$

⇒ $\sf\ (BD)^2 = (AB−BE)^2 + (CE)^2$

$\sf\ [ Since EF=AB ]$

⇒$\sf\ (BD)^2 =(AB)^2 + (BE)^2−2×AB×BE$

$\sf\ +(CE)^2--- (2)$

Adding (1) and (2), we get

$\sf\ (AC)^2 +(BD)^2 =(AB)^2 +(BE)^2$

$\sf\ +2×AB×BE+(CE)^2 +(AB)^2 +(BE)^2−$

$\sf\ 2×AB×BE+(CE)^2$

⇒ $\sf\ (AC)^2 +(BD)^2 =2(AB)^2 +2(BE)^2$

$\sf\ +2(CE)^2$

⇒ $\sf\ (AC)^2 +(BD)^2 =2(AB)^2 +2[(BE)^2 +$

$\sf\ (CE)^2] --- ( 3 )$

In right angled △BEC,

⇒ $\sf\ (BC)^2 =(BE)^2 +(CE)^2$

$\sf\ [ By\: Pythagoras\: theorem ]$

Hence equation ( 3 ) becomes,

⇒$\sf\ (AC)^2 +(BD)^2 =2(AB)^2+2(BC)^2$

⇒$\sf\ (AC)^2 +(BD)^2 =(AB)^2 +(AB)^2+$

$\sf\ (BC)^2+(BC)^2$

∴ $\sf\ (AC)^2 +(BD)^2 =(AB)^2+(BC)^2$

$\sf\ +(CD)^2 +(AD)^2$

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

Answer 2

Answer:

yes it's equal.

Because square change, you get parallelogram..

it's easy...

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