Question

In a triangle $ABC$, $DE$ is parallel to $BC$ if, $\frac{AD}{DB}=\frac{2}{3}$ and $AC=18cm$ find $AE$.

Answer 1

Step 1:Prove that the triangle $ABC$ and the triangle $ADE$ are similar.

Since, $DE$ is parallel to $BC$. and $A$ is the common vertex.

Therefore, $\angle ADE=\angle ABC$ and $\angle AED=\angle ACE$ {Corresponding angles}.

Therefore, the triangle $ABC$ and the triangle $ADE$ are similar by angle-angle similarity.

Step 2: Find the value of $AE$.

Properties of the similar triangle:

1. All the corresponding angles are equal.
2. The ratio of all the corresponding sides is equal.

Therefore,

$\frac{AD}{DB}=\frac{AE}{EC}=\frac{DE}{BC}=\frac{2}{3}$

Assume that, $AE=xcm$. So, $EC=18-xcm$.

Therefore,

$$\begin{gathered} \Rightarrow \frac{x}{18-x}=\frac{2}{3} \\ \Rightarrow 3x=36-2x \\ \Rightarrow 3x+2x=36 \\ \Rightarrow 5x=36 \\ \Rightarrow x=7.2 \end{gathered}$$

Hence, the length of the line segment $AE$ is $7.2cm$.