Question

Please solve
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.

Answer 1

Given,ABCD is a trapezium, AB//DC

DC=2AB,BE:EC=3:4

to prove,7EF=10AB

proof, construction, draw:,AE/BC

DE=BC

DE=BC=2AB(E is the mid point)

In leAEC

AC=AE+EC-1

In le ADE

AD=AE+EC-2

1-2=AC-AD=EC-DE

AC-EC+DE=AD

AB-(2ab) +(BE-BD) =AD

AB-4ab+2bc-4bc=AD

AB-2ab+ab/10=ab

20ab-10ab+ab/10=AF

10AB=7EF