Question

In a rhombus ABCD the diagonals bisect at O prove that AC^2+BD^2=4AB^2

Answer 1

$<b> <I>$
 Hey there !! 

Given :- A rhombus ABCD whose diagonals AC and BD intersect at O.

To Prove :- 4AB² = ( AC² + BD² ).

Proof :- 

➡ We know that the diagonals of a rhombus bisect each other at right angles.


$=> \bf { \angle AOB = \angle BOC = \angle COD = \angle DOA = 90°, }$


$\bf { OA = \frac{1}{2} AC \: and \: OB = \frac{1}{2} BD . }$


From right ∆AOB , we have

AB² = OA² + OB² [ by Pythagoras' theorem ]


$\bf { => {AB}^{2} = (\frac{1}{2} {AC})^{2} + ( \frac{1}{2}{BD})^{2} }.$


$\bf { => {AB}^{2} = \frac{1}{4} ( {AC}^{2} + {BD}^{2} ) }$


=> 4AB² = ( AC² + BD² ).

✔✔ Hence, it is proved ✅✅.

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$\huge \boxed{ \mathbb{THANKS}}$


$\huge \bf{ \# \mathbb{B}e \mathbb{B}rainly.}$

Answer 2

$\huge{hey}$

$\huge \bf{here \: is \: answer}$
$<b> given</b>$

$<b><I> ABCD is a rhombus and diagonal bisect at o ..</b></I>$

$<b><I> to prove :4 AB²= Ac²+BD² </b></I>$

$<b><I> proof : </b></I>$
$<b><I> Ac =1/2OA BD=1/2 OB In ∆AOB </b></I>$


$<b><I> AB²= OA²+OB² AB²=1/2Ac²+1/2BD² AB²=1/4AC²+1/4BD² 4AB²=AC²+BD² proved...</b></I>$

$\huge{ \mathfrak{hope \: it \: helps}}$

$\huge{ \mathbb{Thanks}}$