Question

for a rhombus ABCD, prove that 4AB^2=AC^2+BD^2​

Answer 1

Answer:

Given : A rhombus ABCD.

To prove : $4 {ab}^{2} = {ac}^{2} + {bd}^{2}$

proof : the diagonals of a rhombus bisect each other at right angles.

$oa = \frac{1}{2} ac$

$ob = \frac{1}{2} bd$

$aob = {90}^0{}$

$now \: in \: right \: angle \: triangle \\ aob \: using \: the \: above \: theorem$

${ab}^{2} = {oa}^{2} + {ob}^{2}$

$= ( \frac{1}{2} {ac}^{2} ) + ( \frac{1}{2} {bd}^{2} )$

$= {ac}^{2} + {bd}^{2}$

$4 {ab}^{2} = {ac}^{2} + {bd}^{2}$

=> $hence \: proved$

Step-by-step explanation:

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