Question

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

Answer 1

Answer:

Step-by-step explanation:

Given:

• A rhombus ABCD
• Diagonals AC and BD are perpendicular bisectors

To Prove:

• AB² + BC² + CD² + DA² = BD² + AC²

Proof:

In a rhombus we know that the diagonals are perpendicular bisectors.

Hence,

OD =  OB = 1/2 BC------(1)

OC = OA =  1/2 AC------(2)

Also in a rhombus, all the 4 sides are equal.

Therefore,

AB = BC = CD = DA------(3)

Now consider ΔODC

By Pythagoras theorem,

DC² = OD² + OC²

Substitute value of OD and OC from equations 1 and 2,

DC² = (1/2 BD)² + (1/2 AC)²

DC² = 1/4 BD² + 1/4 AC²

Multiply the whole equation by 4

4DC² = BD² + AC²

DC² + DC² + DC² + DC² = BD² + AC²

Substitute equation 3 in the above equation,

AB² + BC² + CD² + DA² = BD² + AC²

Hence proved.

                              $\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(4,0){2}{\line(0,1){4}}\multiput(0,0)(0,4){2}{\line(1,0){4}}\put(-0.5,-0.5){\bf A}\put(-0.5,4.2){\bf D}\put(4.2,-0.5){\bf B}\put(4.2,4.2){\bf C}\qbezier(0,0)(0,0)(4,4)\qbezier(4,0)(4,0)(0,4)\put(1.8,1.5){\bf{O}}\end{picture}$

Answer 2

Diagram:-

$\setlength{\unitlength}{1cm}\begin{picture}(20,15)\thicklines \put (0,0){\line (1,3){1.5}}\put (0,0){\line (1,0){5}}\put (5,0){\line (1,3){1.5}}\put (1.5,4.5){\line (1,0){5}}\qbezier(1.56,4.5)(1.56,4.5)(5,0)\qbezier(6.45,4.5)(6.45,4.5)(0,0)\put (-0.2,-0.3){\sf B}\put (1.2,4.5){\sf A}\put (5.2,-0.3){\sf C}\put (6.5,4.5){\sf D}\put(3,1.8){\sf O}\end {picture}$

$\Huge \bf \: Construction$

1. Draw a rhombus ABCD.
2. Then mark its centre as O.
3. Draw all diagonal to O

$\huge \bf \: Proof$

As we know that all diagonal are equal in size.

Therefore,

AB = BC = CD = AD

Also,

AC perpendicular to BD

We have to proof

AB² + BC² + CD² + AD² = AC² + BD²

Now ∆AOD we will proof by Pythagoras theorem.

AD² = AO² + OD² (EQ 1)

Similarly,

DC² = DO² + OD² (EQ 2)

BC² = OB² + OC² (EQ 3)

AB² = AO² + BO² (EQ 4)

By adding all equation we will get

AB² + BC² + CD² + AD² = 2AO²+ 2BO² 2CO² + 2DO² (EQ 5)

DO = OB = ½ DB and AO = OC = ½ AC (EQ 6)

Now,

AB² + BC² + CD² + AD²

$\sf { \dfrac{ac}{2} }^{2} + \dfrac{ {bd}^{2} }{2} + \dfrac{ac {}^{2} }{2} + \dfrac{b {d}^{2} }{2}$

Therefore

AB² + BC² + CD² + AD² = BD² + AC²

Hence proved