In line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and $\angle\text{AEF}=\angle\text{AFE}.$ Prove that $\frac{\text{BD}}{\text{CD}}=\frac{\text{BF}}{\text{CE}}.$
[Hint: Take point G on AB such that CG || DF]
[Hint: Take point G on AB such that CG || DF]
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In the given figure of $\triangle\text{ABC},$
EA = AF = EC
EF and BC meets at D.
To Prove: $\frac{\text{BD}}{\text{CD}}=\frac{\text{BF}}{\text{CE}}$
Construction: Draw CG || EF.
Proof: In $\triangle\text{ACG},\text{CG }\parallel\text{ EF}$
E is mid point at AC
F will be the mid point of AG.
⇒ FG = FA
⇒ EC = EA = AF [Given]
⇒ FG = FA = EA = EC .....(i)
In $\triangle\text{BCG}$ and BDF,
CG || EF [by construction]
$\therefore\frac{\text{BC}}{\text{CD}}=\frac{\text{BG}}{\text{GF}}$
$\Rightarrow\frac{\text{BC}}{\text{CD}}+1=\frac{\text{BG}}{\text{GF}}+1$
$\Rightarrow\frac{\text{BC}+{\text{CD}}}{\text{CD}}=\frac{\text{BG}+{\text{CD}}}{\text{GF}}$
$\Rightarrow\frac{\text{BD}}{\text{CD}}=\frac{\text{BF}}{\text{GF}}$
$\Rightarrow\text{FG}={\text{CE}}$ [From (i)]
$\Rightarrow\frac{\text{BD}}{\text{CD}}=\frac{\text{BF}}{\text{CE}}$
Hence, provede.
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