In figure (ii) given below, $A B ∥ D E$ and $B D ∥ E F$ . Prove that $D C^{2} = C F \times A C$ .

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Solution

$△ A B C, \frac{C D}{D A} = \frac{C E}{E B}$ as $A B | | D E$
$△ C D B, \frac{C F}{F D} = \frac{C E}{E D}$ as $E F | | B D$
hence, we get $\frac{C D}{D A} = \frac{C F}{F D}$
Reciprocals $\frac{D A}{C D} = \frac{F D}{C F}$
Add $⊥$ on both sides $\frac{A C}{D C} = \frac{D C}{C F}$
$D C^{2} = C F \times A C$

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