Question

ABCD is a rhombus and its diagonals meet at O, show that: AD²+DC²+CB²+BA²=4DO²+4AO²​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

ABCD is a rhombus and its diagonals AC and BD meet at O.

We know, in rhombus diagonals bisects each other at right angles.

So, AC and BD bisect each other at right angles.

Also, all the sides of rhombus are equal.

Let assume that AB = BC = CD = DA = x

In right-angle triangle AOD

By using Pythagoras Theorem, we have

$\rm \: {AO}^{2} + {DO}^{2} = {x}^{2}$

On multiply by 4, on both sides we get

$\rm \: 4{AO}^{2} + 4{DO}^{2} = 4{x}^{2}$

can be further rewritten as

$\rm \: 4{AO}^{2} + 4{DO}^{2} = {x}^{2} + {x}^{2} + {x}^{2} + {x}^{2}$

can be further rewritten as

$\rm \: 4{AO}^{2} + 4{DO}^{2} = {AB}^{2} + {BC}^{2} + {CD}^{2} + {DA}^{2}$

Hence, Proved

$\rule{190pt}{2pt}$

Theorem Used :-

1. Pythagoras Theorem :-

This theorem states that : In a right-angled triangle, the square of the longest side is equal to sum of the squares of remaining sides.

Additional Information :-

1. Converse of Pythagoras Theorem :-

This theorem states that : If the square of the longest side is equal to sum of the squares of remaining two sides, angle opposite to longest side is right angle.

Answer 2

Step-by-step explanation:

Considerthe given rhombus ABCD.

Let AC & BD be the diagonals of rhombus ABCD

In Δ A O D by phythagoras theorem,

$\\ \tt \[ A B^{2}=A O^{2}+D O^{2} \]$

By multiplying by 4 on both side and we get,

$\\ \tt \[ 4 A B^{2}=4 A O^{2}+4 D O^{2} \]$

We know that, the all sides of rhombus are equal Then,

$\\ \tt A B=B C=C D=D A$

So,

$\tt A B^{2}+B C^{2}+C D^{2}+D A^{2}$

$\\ \tt=4 A O^{2}+4 D O^{2}$

Hence proved.