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) = (ABCD)

But in quadrilateral EBGD, we have

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

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

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

(EHG) = (ABCD) - (EHG) or

2(EHG) = (ABCD) or

(EHG) / (ABCD) = 1/4