In the figure, $\triangle ABC$ and $\triangle CEF$ are two triangles where $BA$ is parallel to $CE$ and $\dfrac{AF}{AC} = \dfrac{5}{8}$.
- Prove that $\;$ $\triangle ADF \sim \triangle CEF$.
- Find $AD$ if $CE = 6 \; cm$.
- If $DF \parallel BC$, find $\;$ $\dfrac{\text{Area} \left(\triangle ADF\right)}{\text{Area} \left(\square DFCB\right)}$.
- Given: $\;$ $BA \parallel CE$
$\implies$ $AD \parallel CE$ $\;\;\;$ [$B - D - A$ are collinear points]
$AC$ is a transversal.
$\therefore \;$ $\angle DAF = \angle FCE$ $\;\;\;$ [alternate angles]
$\angle AFD = \angle CFE$ $\;\;\;$ [vertically opposite angles]
$\therefore \;$ by AA (angle - angle) axiom, $\;$ $\triangle ADF \sim \triangle CEF$ - $\because \;$ $\triangle ADF \sim \triangle CEF$,
$\dfrac{AD}{CE} = \dfrac{AF}{CF}$ $\;\;\; \cdots \; (1)$ [corresponding sides of similar triangles are in proportion]
Given: $\;$ $\dfrac{AF}{AC} = \dfrac{5}{8}$
i.e. $\;$ $\dfrac{AF}{AF + CF} = \dfrac{5}{8}$
i.e. $\;$ $\dfrac{AF + CF}{AF} = \dfrac{8}{5}$
i.e. $\;$ $1 + \dfrac{CF}{AF} = \dfrac{8}{5}$
i.e. $\;$ $\dfrac{CF}{AF} = \dfrac{8}{5} - 1 = \dfrac{3}{5}$
$\therefore \;$ $\dfrac{AF}{CF} = \dfrac{5}{3}$ $\;\;\; \cdots \; (2)$
In view of equation $(2)$, equation $(1)$ becomes
$\dfrac{AD}{CE} = \dfrac{5}{3}$
i.e. $\;$ $AD = CE \times \dfrac{5}{3} = 6 \times \dfrac{5}{3} = 10 \; cm$ - $\because \;$ $DF \parallel BC$ and $AB$ is the transversal,
in triangles $ADF$ and $ABC$,
$\angle ADF = \angle ABC$ $\;\;\;$ [corresponding angles]
$\angle DAF = \angle BAC$ $\;\;\;$ [common angle]
$\therefore \;$ $\triangle ADF \sim \triangle ABC$ $\;\;\;$ [by A.A axiom]
Now, ratio between the area of two similar triangles is equal to the ratio between the squares of its corresponding sides.
$\therefore \;$ $\dfrac{\text{Area} \left(\triangle ADF\right)}{\text{Area} \left(\triangle ABC\right)} = \dfrac{AF^2}{AC^2} = \dfrac{5^2}{8^2} = \dfrac{25}{64}$
i.e. $\;$ $\dfrac{\text{Area} \left(\triangle ABC\right)}{\text{Area} \left(\triangle ADF\right)} = \dfrac{64}{25}$ $\;\;\; \cdots \; (3)$
Now, $\;$ $\dfrac{\text{Area} \left(\triangle ADF\right)}{\text{Area} \left(\square DFCB\right)} = \dfrac{\text{Area} \left(\triangle ADF\right)}{\text{Area} \left(\triangle ABC\right) - \text{Area} \left(\triangle ADF\right)}$
$\begin{aligned} i.e. \; \dfrac{\text{Area} \left(\square DFCB\right)}{\text{Area} \left(\triangle ADF\right)} & = \dfrac{\text{Area} \left(\triangle ABC\right) - \text{Area} \left(\triangle ADF\right)}{\text{Area} \left(\triangle ADF\right)} \\\\ & = \dfrac{\text{Area} \left(\triangle ABC\right)}{\text{Area} \left(\triangle ADF\right)} - 1 \\\\ & = \dfrac{64}{25} - 1 \;\;\; [\text{in view of equation (3)}] \\\\ & = \dfrac{39}{25} \end{aligned}$
$\therefore \;$ $\dfrac{\text{Area} \left(\triangle ADF\right)}{\text{Area} \left(\square DFCB\right)} = \dfrac{25}{39}$