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?

1 Answer
May 10, 2018

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.