Question
In Fig. $D E \| B C$ and $C D \| E F$. Prove that $AD ^2= AB \times AF$.
✓
Answer
In $\triangle ABC$, it is given that
$D E \| B C$
$\Rightarrow \frac{A B}{A D}=\frac{A C}{A E} .........(i)$
In $\triangle ADC$, it is given that
$F E \| D C$
$\Rightarrow \frac{A D}{A F}=\frac{A C}{A E} .........(ii)$
From $(i)$ and $(ii),$ we get
$\frac{A B}{A D}=\frac{A D}{A F}$
$\Rightarrow AD^2=AB \times AF$
$D E \| B C$
$\Rightarrow \frac{A B}{A D}=\frac{A C}{A E} .........(i)$
In $\triangle ADC$, it is given that
$F E \| D C$
$\Rightarrow \frac{A D}{A F}=\frac{A C}{A E} .........(ii)$
From $(i)$ and $(ii),$ we get
$\frac{A B}{A D}=\frac{A D}{A F}$
$\Rightarrow AD^2=AB \times AF$
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