Question

In $\Delta ABC$, AD is median on BC, E is on AD such that BE = AC. If the line BE intersects AC at F when produced, then AF is always equal to

• A. AE
• B. ED
• C. FE
• D. FC

Answer 1

The correct option is C FE

From C, draw CG || AD intersecting BF when produced at G.


In $\Delta EAF$ and $\Delta GCF$, $\angle EAF=\angle FCG$ (Alternate interior angles)
$\angle AEF=\angle FGC$ (Alternate interior angles)
$\therefore \Delta EAF\sim \Delta GCF$ (AA Similarity)
$\Rightarrow \frac{AF}{CF}=\frac{EF}{GF}\Rightarrow \frac{AF}{EF}=\frac{CF}{GF}$
Let $\frac{AF}{EF}=\frac{CF}{GF}=k$
Then, $AF=k\left(EF\right)$ and $CF=k\left(GF\right)$
So, $\frac{AF+CF}{EF+GF}=\frac{k\left(EF\right)+k\left(GF\right)}{EF+GF}=k$
$\therefore \frac{AF}{EF}=\frac{CF}{GF}=\frac{AF+CF}{EF+GF}=\frac{AG}{EG}$ .......(i)
In $\Delta BGC,DE||CG$
So, mid point theorem, $\frac{BD}{DC}=\frac{BE}{EG}$
Since $BD=DC\Rightarrow BE=EG$
$\Rightarrow EG=AC(BE=AC)$......(ii)
From (i) and (ii), $AF=FE$
Hence the correct answer is option (3)