The problem is: The area of a rhombus is one-half the product of 
the lengths of its diagonals. 
I know the answer as with the previous problem, I just do not know 
how to go about writing a proof for this. 
Thank You

I won't do it in numbered steps and reasons.  You can do that. I'll 
just explain what you do to prove it.


    A                     Given: Rhombus ABCD
   /|\                    To prove: Area of Rhombus ABCD = DB×AC/2
  / | \
D/ E|  \B
 \¯¯|¯¯/
  \ | /
   \|/ 
   C

Area of rhombus ABCD = area of triangle ABD + area of 
triangle CBD

Triangles ABD and CBD are congruent by SSS

Area of rhombus ABCD = 2×(Area of triangle ABD)

AE is perpendicular to DB because the diagonals 
of a rhombus are perpendicular bisectors of each other.

Area of triangle ABD = DB×AE/2 because a triangle's area
is one-half the product of a side and the altitude drawn
to that side.

Area of rhombus ABCD = 2×(Area of triangle ABD)

So area of rhombus ABCD = 2×(DB×AE/2) = DB×AE

AE = AC/2 because the diagonals of a rhombus are perpendicular
bisectors of each other.

So area of rhombus DB×AE = DB×(AC/2) = DB×AC/2

Edwin