Question

ABCD is a rhombus. AC and BD are its two diagonals intersecting at O. Prove that AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2?

Answer 1

Please see the explanation.

Explanation:

The sides of rhombus are congruent by definition, the diagonals that are perpendicular bisectors divide the figure into 4 congruent right triangles, se we can conclude the followings:

`DeltaOAB~=DeltaOBC~=DeltaOCD~=DeltaOAD`

`AB^2=OA^2+OB^2`
`BC^2=OB^2+OC^2`
`CD^2=OC^2+OD^2`
`DA^2=OA^2+OD^2`
But:
`AC=OA+OC`
`AC^2=(OA+OC)^2=OA^2+2*OA*OC+OC^2`
`OA=OC`
`AC^2=2*OA^2+2*OC^2`
Also:
`BD=OB+OD`
`BD^2=2*OB^2+2*OD^2`
So:
`AB^2+BC^2+CD^2+DA^2=2*OA^2+2*OB^2+2*OC^2+2*OD^2`
Thus:
`AB^2+BC^2+CD^2+DA^2=AC^2+BD^2`

Hence proved.