Trapezoid bases are treated as 2: 3. A straight line parallel to the bases is drawn through the intersection of the diagonals. In what respect does this line divide the area of the trapezoid?
By the ninth property of the trapezoid, the triangles AOD and BOC are similar.
Therefore, BC / AD = OC / AO = 2/3
Draw a line perpendicular to the bases through the point of intersection of the diagonals.
Consider the triangles AOF and COE.
∠OAF = ∠OCE (crosswise-lying angles).
∠AFO = ∠CEO = 90 °
Therefore, these triangles are similar (by the first sign of the similarity of triangles).
Then, OC / AO = OE / OF = 2/3
For simplicity, denote BC as 2x and AD as 3x
According to the fifth trapezoid property, GH = 2 * 2x * 3x / (2x + 3x) = 12×2 / 5x = 12x / 5
Area of the upper trapezoid:
S1 = (BC + GH) * EO / 2 = (2x + 12x / 5) * EO / 2 = (10x + 12x) * EO / 10 = 22x * EO / 10
Area of the lower trapezoid:
S2 = (AD + GH) * OF / 2 = (3x + 12x / 5) * OF / 2 = (15x + 12x) * OF / 10 = 27x * OF / 10
S1 / S2 = (22x * EO / 10) / (27x * EO / 10) = (22x * EO) / (27x * OF) = 22EO / 27OF = 22 * 2 / (27 * 3) = 44/81
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