Question

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

Answer 1

see explanation.

Explanation:

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`