Question

In figure, line segment DF intersects the side AC of a $\Delta ABC$ at the point E such that E is the mid – point of CA and $\angle AEF$ and $\angle AFE$. Prove that $\frac{BD}{CD}=\frac{BF}{CE}$

Answer 1

Given, $\Delta ABC$, E is the mid-point of CA and $\angle AEF=\angle AFE$
To prove $\frac{BD}{CD}=\frac{BF}{CE}$
Construction Take a point G on AB such that CG II EF.
Proof Since, E is the mid – point of CA.
$\therefore$ CE = AE ………(i)
In $\Delta ACG$, CG II EF and E is mid – point of CA.
So, CE = GF ……(ii)
[ by mid-point theoreum]
Now, in $\Delta BCG$, and $\Delta BDF$, CG II EF
$\therefore$ $\frac{BC}{CD}=\frac{BG}{GF}$ [by basic proportionally therorem]
$\Rightarrow \frac{BC}{CD}=\frac{BF-GF}{GF}\Rightarrow \frac{BC}{CD}=\frac{BF}{GF}-1$
$\Rightarrow \frac{BC}{CD}+1=\frac{BF}{CE}$ [from Eq.(ii)]
$\Rightarrow \frac{BC+CD}{CD}=\frac{BF}{CE}\Rightarrow \frac{BD}{CD}=\frac{BF}{CE}$