Question

In the given figure, $A–D–C$ and $B–E–C$ , $DE\parallel AB$.

If $AD=5,DC=3,BC=6.4$ then Find $BE$.

Answer 1

Given:

In figure,
$AD=5,DC=3,BC=6.4$ and $BE=x$

In $△ABC,DE\parallel AB$

$\Rightarrow \frac{CD}{DA}=\frac{CE}{EB}$ [By basic proportionality theorem]

$\Rightarrow \frac{3}{5}=\frac{6.4-x}{x}$

$\Rightarrow 3x=5(6.4-x)$

$\Rightarrow 3x=32-5x$

$\Rightarrow 3x+5x=32$

$\Rightarrow 8x=32$

$\Rightarrow x=\frac{32}{8}$

$\Rightarrow x=4$

Therefore, $BE=4$