In a trapezium ABCD, AB is parallel to DC and DC = 2AB E is point on BC such - Maths - Triangles

Question

In a trapezium ABCD, AB is parallel to DC and DC = 2AB E is
point on BC such that BE : EC = 3:4 EF is drawn parallel to AB
where F is a point on AD Prove that 7EF = 10AB

Ajanta Trivedi · Accepted Answer

ABCD is a trapezium where $A B ∥ D C$ and DC = 2 AB (1) and BE : EC = 3:4 (2) from eq(2): $\frac{B E}{E C} = \frac{3}{4} \frac{B E}{E C} + 1 = \frac{3}{4} + 1 \frac{B E + E C}{E C} = \frac{3 + 4}{4} \frac{B C}{E C} = \frac{7}{4}$ the triangles CGE and triangles CAB are similar triangles therefore ratio of corresponding sides are equal $\frac{E G}{A B} = \frac{E C}{B C} \frac{E G}{A B} = \frac{4}{7} \Rightarrow 7 E G = 4 A B (3)$ since $A B ∥ E F ∥ C D$ , therefore $\frac{A F}{F D} = \frac{B E}{E C} = \frac{3}{4} \frac{F D}{A F} = \frac{4}{3} \frac{F D}{A F} + 1 = \frac{4}{3} + 1 \frac{F D + A F}{A F} = \frac{4 + 3}{3} \frac{A D}{A F} = \frac{7}{3}$ the triangles AFG and ADC are similar triangles therefore $\frac{F G}{C D} = \frac{A F}{A D}$ $\frac{F G}{C D} = \frac{3}{7} \Rightarrow 7 F G = 3 C D$ from eq(1): $7 F G = 3 * 2 A B 7 F G = 6 A B (4)$ adding eq(3) and eq(4): $7 E G + 7 F G = 4 A B + 6 A B 7 (E G + F G) = 10 A B 7 E F = 10 A B$ which is the required result hope this helps you cheers!!

In a trapezium ABCD, AB is parallel to DC and DC = 2AB E is point on BC such that BE : EC = 3:4. EF is drawn parallel to AB where F is a point on AD . Prove that 7EF = 10AB.