**Problem 4 of the Canadian Mathematical Olympiad 1978**

The sides AD and BC of a convex quadrilateral ABCD are extended to meet at E. Let H and G be the midpoints of BD and AC, respectively. Find the ratio of the area of the triangle EHG to that of the quadrilateral ABCD.

**Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to my sister Nguyễn Thị Hạnh)**

Let (Ω) denote the area of shape Ω. Since H is the midpoint of BD,

(EHD) = (EHB) and (GHD) = (GHB) so that

(EHG) + (EBG) = (EHD) + (GHD) (*)

Similarly, since G is the midpoint of AC,

(ABD) + (GBD) = (BGC) + (DGC) and

(ABD) + (GBD) = (ABGD) = ½ (ABCD)

But in quadrilateral EBGD, we have

(EHG) = ½ (ABCD) + (EAB) – (EBG) – [ (EHD) + (GHD) ]

Substituting (EHD) + (GHD) from (*), we have

(EHG) = ½ (ABCD) + (EAB) – (EBG) - (EHG) - (EBG) (**)

But again, since G is the midpoint of AC, the altitude from G to EB is half the altitude from A to EB, meaning

(EAB) = 2(EGB).

Equation (**) becomes

(EHG) = ½ (ABCD) + 2(EBG) – (EBG) - (EHG) - (EBG) or

(EHG) = ½ (ABCD) - (EHG) or

2(EHG) = ½ (ABCD) or

(EHG) / (ABCD) = 1/4