1) the proportional function y=k 1x, and the inverse proportional function y = k2 \ x.

Let B(p, q)D(q, p)

Bring in K 1 = 1 and k2 =1respectively (both p and q are eliminated).

Bring it in again to get p=- 1 q= 1.

Get b (- 1, 1) d (1, 1).

BA = 1 AO = 1 OC = 1 CD = 1

ABCD is a parallelogram.

Sabcd=AB×AC= 1×2=2

The quadrilateral ABCD is 2.

2) E(- 1，- 1) BE=2 ED=2。

Get s △ bed = de be/2 = 2 * 2/2 = 2.

It is concluded that the areas of quadrilateral ABCD and △DBE are equal.