CDS Model Paper 1 Mathematics

In Δ‌ABC, we draw a line l||BF which intersect AC at G. In Δ‌ADG and Δ‌AEF
given that EA is the mid point of AD and DL||EF.
So, concept of similar triangle.
F is also mid point of AG.
AF=FG ...(i)
Δ‌ADG and Δ‌AEF are similar.
Again
Δ‌FBC and Δ‌DCL
BF||DG
given that AD is median so that CD the mid point of BC. G will be The mid point of CF
CG=GF ...(ii)
From equations (i) and (ii), we get
AF=FG=CG ...(iii)
From figure, AC=AF+FG+CG
=AF+AF+AF+3AF
⇒ AF=

1

3

AC