In which of the following figure, the triangle formed by the line segment DE using the sides of the triangle ABC is similar to the triangle ABC?

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Solution

The correct option is C

According to basic proportionality theorem, if DE || AB, then:
$\frac{A D}{C D} = \frac{B E}{C E}$
$\frac{A D}{C D} + 1 = \frac{B E}{C E} + 1$
$\frac{A D + C D}{C D} = \frac{B E + C E}{C E}$
$\frac{A C}{C D} = \frac{B C}{C E}$

$∠ C$ is common
$∠ C D E = ∠ C A B$ as they are corresponding angles.
$∠ D E C = ∠ A B C$ as they are corresponding angles.

So, $△ A B C \sim △ D E C$ .

In the given options, DE||AB in only one the option shown above. This theorem will hence, hold only for this option.

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