Question

Diagonals of rhombus ABCD intersect each other at point O. Prove that OA square +OC square= 2AD square-BD square by 2

Answer 1

Please see the attachment

Answer 2

OA square +OC square= 2AD square-BD square by 2

Proved below.

Step-by-step explanation:

Given:

Here diagonals of rhombus ABCD intersect each other at point O [as shown in the figure below]

Here AB =  BC = CD = AD             [1]  [ ABCD is a rhombus ]

And we know diagonals of rhombus are perpendicular bisectors of each other , So

$OA = OC = \frac{AC}{2}$                      [2]  

$OB =OD=\frac{BD}{2}$                       [3]

And ∠ AOB = ∠ BOC = ∠ COD = ∠ DOA  = 90°

Now we apply Pythagoras theorem in Δ∆ AOD and get,

$AD^2 = OA^2 + OD^2$                           [4]

And

Now we apply Pythagoras theorem in Δ∆ COD and get ,

$CD^2 = OC^2 + OD^2$,   We substitute value from equation 1 ( CD = AD ) we get

$AD^2 = OC^2 + OD^2$                             [5]

Now we add equation 4 and 5 we get

$2 AD^2 =OA^2 + OC^2 +2 OD^2$  [We substitute value from equation 3 $(OD=\frac{BD}{2} )$ ] we get,

$2 AD^2 = OA^2 + OC^2 + 2(\frac{BD}{2} )^{2}$

$2 AD^2 = OA^2 + OC^2 + 2\times\frac{BD^2}{4}$

$2AD^2 = OA^2 + OC^2 +\frac{BD^2}{2}$

$OA^2 + OC^2 = 2 AD^2 -\frac{BD^2}{2}$

Hence proved.