MCQ
We have, $AB\|DE$ and $BD \| EF$ .Then
- A$B C^2=A B \cdot C E$
- B$A C^2=B C \cdot D C$
- C$A B^2=A C \cdot D E$
- ✓$D C^2=C F \times A C$
✓
Answer
Correct option: D.
$D C^2=C F \times A C$
In $\triangle ABC$, using Thales theorem,
$\frac{DC}{AC}=\frac{CE}{BC}[A B \| D E]\ldots(i)$
And in triangle BCD, using Thales theorem,
$\frac{CF}{DC}=\frac{CE}{BC}[B D \| E F]\ldots(ii)$
From eq. $(i)$ and $(ii)$, we have
$\frac{DC}{AC}=\frac{CF}{DC}$
$\Rightarrow D C^2=C F \times A C$
$\frac{DC}{AC}=\frac{CE}{BC}[A B \| D E]\ldots(i)$
And in triangle BCD, using Thales theorem,
$\frac{CF}{DC}=\frac{CE}{BC}[B D \| E F]\ldots(ii)$
From eq. $(i)$ and $(ii)$, we have
$\frac{DC}{AC}=\frac{CF}{DC}$
$\Rightarrow D C^2=C F \times A C$
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