# Trapezoid bases are treated as 2: 3. A straight line parallel to the bases is drawn through the intersection of the diagonals

**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

Answer: 44/81