Since the area of a triangle is bh/2 and the area of a parallelogram is bh, we have to have bh/2=bh. Since it clearly doesn't work if b and h are the same for both, we'll write the parallelogram's b as x and the parallelogram's h as y, getting bh/2=xy. If the bases are the same, we can cross b and x out, getting h/2=y, meaning that the height of the triangle must be twice the height of the parallelogram. In addition, if the heights are the same, the base of the triangle must be twice the base of the parallelogram by cancelling them out like we did with the bases.
