Question

In $\Delta ABC$, AC = 15 cm and DE || BC. If $\frac{AB}{AD}=3$, find EC.

(3 Marks)

Answer 1

Given: DE$\parallel$BC
From basic proportionality theorem
$\frac{AD}{DB}=\frac{AE}{EC}$ (1 Mark)
$\text{Now,}\frac{AB}{AD}=\frac{AD+DB}{AD}=1+\frac{DB}{AD}\Rightarrow 1+\frac{DB}{AD}=3\Rightarrow \frac{DB}{AD}=2\Rightarrow \frac{AD}{DB}=\frac{AE}{EC}=\frac{1}{2}\Rightarrow 2AE=EC\Rightarrow AC=AE+EC-------\left(1\right)$
On substituting value of EC in (1), we get
$15=3AE\Rightarrow 5=AE\Rightarrow EC=10cm$
(2 Marks)