Question

In the given figure, line segment BE intersect the side CD of a triangle ACD at F. If DF : CF = 1 : 2, then which of the following is always true for $\frac{AB}{BD}$?

• A. $\frac{3AE}{2CE}$
  
• B. $\frac{2AE}{CE}$
  
• C. $\frac{3AE}{CE}$
  
• D. $\frac{2AE}{3CE}$
  

Answer 1

The correct option is B $\frac{2AE}{CE}$



$DF:CF=1:2$
$\Rightarrow \frac{DF}{CF}=\frac{1}{2}$
$\Rightarrow \frac{DF}{CF}+1=\frac{1}{2}+1$
$\Rightarrow \frac{DF+CF}{CF}=\frac{1+2}{2}$
$\Rightarrow \frac{CD}{CF}=\frac{3}{2}$ ......(i)
Construct DG || BE
Thus, $\Delta ABE\sim \Delta ADG$
$\Rightarrow \frac{AB}{BD}=\frac{AE}{GE}$ (By BPT)
Also DG || FE
$\Rightarrow \frac{CG}{CE}=\frac{CD}{CF}$ (By BPT)
$\Rightarrow \frac{CG}{CE}=\frac{3}{2}$
$\Rightarrow CG=\frac{3}{2}CE$ ......(ii)
$\therefore GE=CG-CE\frac{1}{2}CE$ [from (ii)]
$\Rightarrow \frac{AB}{BD}=\frac{AE}{\frac{1}{2}CE}=\frac{2AE}{CE}$
Hence, the correct answer is option (2).