Question

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in a rhombus ABC. prove that

AC^2 +BD^2=4BC^2

Please help me

Thankyou

Answer 1

rhombus is of 4 sides ABCD what you had written ABC and the answer is1) The sides of a rhombus are all congruent (the same length.) AB=BC=CD=DA=a

2) Opposite angles of a rhombus are congruent (the same size and measure.)
∠BAD=∠BCD=y,and∠ABC=∠ADC=x

3) The intersection of the diagonals of a rhombus form 90 degree (right) angles. This means that they are perpendicular. 
∠AOB=∠BOC=∠COD=∠DOA=90∘

4) The diagonals of a rhombus bisect each other. This means that they cut each other in half.
BO=OD=12BD=m,andAO=OC=12AC=n

5) Adjacent sides of a rhombus are supplementary. This means that their measures add up to 180 degrees.
x+y=180∘

Now back to our question.

In ΔBOC,BC2=BO2+OC2
Since BO=12BD,andOC=12AC

⇒BC2=(12BD)2+(12AC)2

⇒B
hope it is helpful

Answer 2

$\huge \boxed{ \mathfrak{Hello \: Friend }}$

$To \: prove:- \\ \\ {AC}^{2} +{BD}^{2} ={4BC}^{2}$

$Proof \colon -$

$In \: \triangle \: BOC, \: By \: Pythagoras \\ Theorem$

${BC}^{2} = {BO}^{2} + {OC}^{2}$

$Since \: BO ={ ( \frac{1}{2} BD) } \: and \: \\ OC = {(\frac{1}{2}AC)}$

$\Rightarrow {BC}^{2} ={ (\frac{1}{2} BD)}^{2} + ( { \frac{1}{2}AC})^{2}$

$\Rightarrow {BC}^{2} = \frac{1}{4} {( BD)}^{2} + \frac{1}{4} {(AC )}^{2}$

$\Rightarrow{ BC}^{2} = \frac{1}{4} ( {BD}^{2} +{ (AC)}^{2}$

$\Rightarrow {(4BC)}^{2} ={ BD}^{2} + {AC}^{2}$

$\mathcal{Hope \: the \: above \: answer \: helped }$

$\mathbb{PLEASE \: MARK \: BRAINLIEST}$
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