Question
In figure $DE \| AC$ and $DF \| AE$. Prove that $\frac{B F}{F E}=\frac{B E}{E C}$
✓
Answer
In $\triangle A B E$, we have $D F \| A E$, then
$\frac{B D}{A D}=\frac{B F}{F E}[$ By BPT $] \ldots \ldots$ (i)
In $\triangle A B C$, we have $D E \| A C$, then
$\frac{B D}{A D}=\frac{B E}{E C}[$ By BPT $] \ldots \ldots$ (2)
From (i) and (2), We get
$\frac{B F}{F E}=\frac{B E}{E C}$
$\frac{B D}{A D}=\frac{B F}{F E}[$ By BPT $] \ldots \ldots$ (i)
In $\triangle A B C$, we have $D E \| A C$, then
$\frac{B D}{A D}=\frac{B E}{E C}[$ By BPT $] \ldots \ldots$ (2)
From (i) and (2), We get
$\frac{B F}{F E}=\frac{B E}{E C}$
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