Question

Prove that the sum of the squares of the sides of a rhombus equal to the sum
of the squares of its diagonals​

Answer 1

$AC 2 =OA 2 +OC 2 +2AO.OC=4AO 2$

□ABCD is a rhombus with O as point of intersection of diagonals.

In ΔAOB,

∠AOB=90

0 (since diagonals are perpendicular in rhombus).

By Pythagoras theorem,

AB ^2 =AO ^2 +OB ^2

Similarly,

BC^2 =OC^2 +OB ^2 ,DC ^2 =OD^ 2+OC ^2

DA ^2 =DO ^2 +OA ^2

AB^2 +BC ^2 +CD ^2 +DA ^2 =2(OA ^2 +OB ^2 +OC ^2 +OD ^2 =4(AO^ 2 +DO^ 2 )  

Rhombus diagonal biset each other,

AO=OC,DO=OB

AC=AO+OC

AC 2 =OA^2 +OC^2 +2AO.OC=4AO^2

Similarly,

DB 2=4OD

∴AC^2 +DB ^2 =4(AO^2 +DO^2 )

AB^2 +BC^2 +CD^2 +DA^2 =AC^2 +DB^2

Hence Proved.

Answer 2

Given :

• ◼️ABCD is a parallelogram.
• Diagonals AC and BD intersects at point O.

To prove :

• AC² + BD² = AB² + BC² + CD² + AD²

Proof :

Since ◼️ABCD is a parallelogram,

•°• CD = AB (1)

AD = BC (2)

$\sf{\big[Opposite\:sides\:of\:a\:parallelogram\:are\:equal\big]}$

Now, we know that the diagonals of a parallelogram bisect each other.

•°• AO = OC = $\sf{\dfrac{1}{2}\:\times\:AC\:\:\:(3)}$

Also, BO = OD = $\sf{\dfrac{1}{2}\:\times\:BD\:\:\:(4)}$

In ΔABC, we know that seg BO is the median.

The relation between the median and the sides of a triangle is given using Apollonius theorem.

•°• by Apollonius thm,

$\sf{AB^2 + BC^2\:=\:2\:BO^2\:+\:2\:OC^2}$

Now, using equations (3) and (4), we can replace LHS,

$\sf{AB^2\:+\:BC^2\:=\:2\:\Big(\dfrac{1}{2}\:\times\:BD\:\Big)^2\:+\:2\:\Big(\dfrac{1}{2}\:\times\:AC\:\Big)^2}$

$\sf{AB^2\:+\:BC^2\:=\:\cancel{2}\:\times\:\dfrac{1}{\cancel{4}}\:\times\:BD^2\:+\:\cancel{2}\:\times\:\dfrac{1}{\cancel{4}}\:\times\:AC^2}$

$\sf{AB^2\:+\:BC^2\:=\:\dfrac{1}{2}\:\times\:BD^2\:+\:\dfrac{1}{2}\:\times\:AC^2}$

Now, multiply throughout by 2,

$\sf{2\:\times\:AB^2\:+\:2\:\times\:BC^2\:=\:\cancel{2}\:\times\:\dfrac{BD^2}{\cancel{2}}\:+\:\cancel{2}\:\times\:\dfrac{AC^2}{\cancel{2}}}$

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

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

From equations (1) and (2), we can replace AB and BC,

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

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

$\large{\boxed{\bold{AC^2+BD^2=AB^2+BC^2+CD^2+AD^2}}}$