In figure below, line segment $ \mathrm{DF} $ intersect the side $ \mathrm{AC} $ of a triangle $ \mathrm{ABC} $ at the point $ \mathrm{E} $ such that $ \mathrm{E} $ is the mid-point of $ \mathrm{CA} $ and $ \angle \mathrm{AEF}=\angle \mathrm{AFE} $. Prove that $ \frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{BF}}{\mathrm{CE}} $
[Hint: Take point $ \mathrm{G} $ on $ \mathrm{AB} $ such that $ \mathrm{CG} \| \mathrm{DF} $.]
"
Given:
Line segment \( \mathrm{DF} \) intersect the side \( \mathrm{AC} \) of a triangle \( \mathrm{ABC} \) at the point \( \mathrm{E} \) such that \( \mathrm{E} \) is the mid-point of \( \mathrm{CA} \) and \( \angle \mathrm{AEF}=\angle \mathrm{AFE} \).
To do:
We have to prove that \( \frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{BF}}{\mathrm{CE}} \)
Solution:
Take a point $G$ on $A B$ such that $C G \| E F$.
$E$ is the mid-point of $C A$
This implies,
$C E=A E$..........(i)
In $\triangle A C G, C G \| E F$
$E$ is the mid-point of $C A$.
This implies,
$C E=G F$........(ii)
In $\triangle B C G$ and $\triangle B D F, C G \| E F$
By using basic proportionality theorem, we get,
$\frac{B C}{C D}=\frac{B G}{G F}$
$\frac{B C}{C D}=\frac{B F-G F}{G F}$
$\frac{B C}{C D}=\frac{B F}{G F}-1$
$\frac{B C}{C D}+1=\frac{B F}{C E}$ [From (ii)]
$\frac{B C+C D}{C D}=\frac{B F}{C E}$
$\frac{B D}{C D}=\frac{B F}{C E}$
Hence proved.
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