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# 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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Updated on: 10-Oct-2022

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