If #ABCD# is a rhombus, how do you prove that #4BC^2=AC^2+BD^2#?

Rhombus properties :

1) 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.)
#angleBAD=angleBCD=y, and angleABC=angleADC=x#

3) The intersection of the diagonals of a rhombus form 90 degree (right) angles. This means that they are perpendicular.
#angleAOB=angleBOC=angleCOD=angleDOA=90^@#

4) The diagonals of a rhombus bisect each other. This means that they cut each other in half.
#BO=OD=1/2BD=m, and AO=OC=1/2AC=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 #DeltaBOC, BC^2=BO^2+OC^2#
Since #BO=1/2BD, and OC=1/2AC#

#=> BC^2=(1/2BD)^2+(1/2AC)^2#

#=> BC^2=1/4(BD)^2+1/4(AC)^2#

#=> BC^2=1/4(BD^2+AC^2)#

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