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AAA similarity

Refer to the ...

Question

Refer to the figure: DE is parallel to BC.

$\frac{A D}{D B}$ = ?

A

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B

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C

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D

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Solution

The correct option is A

Consider $△ A D E$ and $△ A B C$

$∠ D A E = ∠ B A C$ [Common angle]

$∠ A D E = ∠ A B C$ [right angle]

Therefore by AA similarity criterion, the triangles $△ A D E$ and $△ A B C$ are similar.

So, $\frac{A D}{A B}$ = $\frac{A E}{A C}$

$⟹$ $\frac{A D}{A D + D B}$ = $\frac{A E}{A E + E C}$

$⟹$ $\frac{A D + D B}{A D}$ = $\frac{A E + E C}{A E}$

$⟹$ 1 + $\frac{D B}{A D}$ = 1 + $\frac{E C}{A E}$

$⟹$ $\frac{D B}{A D}$ = $\frac{E C}{A E}$

$⟹$ $\frac{A D}{D B}$ = $\frac{A E}{E C}$

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