Question

In a trapezium ABCD, AB is parallel to DC and DC=3AB. A line EF is drawn parallel to AB and DC such that it cuts AD at E and BC at F and BF/ FC = 2 / 3. Prove that : 3DC=5EF.

Answer 1

In a trapezium ABCD, AB is parallel to DC and DC = 3 AB.

Let AB be x.
DC = 3x

Consider ΔBCD and ΔBOF,

∠BOF = ∠BDC, (Corresponding angles)
∠BFO = ∠BCD, (Corresponding angles)
∠OBF = ∠OBF, (Common angles)

ΔBCD ~ ΔBOF (AA similarity criterion)

BF/FC = BO/BD = OF/DC = 2/5 [Since BF/ FC = 2 / 3, adding 1 on both sides]
OF/DC = 2/5

OF = 2/5 × 3x = 6x / 5

Similarly, ΔADB ~ ΔEDO (AA similarity criterion)

BO/BD = AE/AD = OE/AB = 3/5

OE/AB = 3/5

OE = 3/5 × x

3 DC = 5 EF

3 DC = 3 × 3x = 9x

EF = OE + FO = 3x / 5 + 6x / 5 = 9x / 5

5 EF = 5 (9x / 5) = 9x

∴ 3 DC = 5 EF