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Tiny But Not So Simple Geometry Problem : Areas From Quadrilateral Diagonals

Question.A convex quadrilateral is divided into four parts by its diagonals. Three of the areas are 2, 1, and 3 as shown in the diagram. What is the area of the fourth region denoted by x?

solution to above question:

Let’s start solve a more general problem of the four regions as s₁, s₂, s₃, s₄.
Label the quadrilateral with vertices A, B, C, D and write E for the point where the diagonals intersect. Now draw an altitude from A to BD, with length h₁, and another altitude from C to BD, with length h₂.
The areas of each region can be found using the formula for the area of a triangle as shown in figure.

where,
s₁ = (DE)h₁/2 s₂ = (BE)h₁/2 s₃ = (BE)h₂/2 s₄ = (DE)h₂/2
Thus we have:
s₁(s₃) = s₂(s₄)
The original problem had s₁ = 2, s₂ = 1, s₃ = 3, and s₄ = x. So we have:
(2)(3) = (1)x
6 = x