In the figure, $ABC$ and $CEF$ are two triangles where $BA$ is parallel to $CE$ and $AF : AC = 5 : 8$.
- Prove that $\triangle ADF \sim \triangle CEF$
- Find $AD$ if $CE = 6 \; cm$.
- If $DE \parallel BC$, find $\;$ $\dfrac{\text{Area } \left(\triangle ADF\right)}{\text{Area } \left(\triangle ABC\right)}$.
- $BA \parallel CE$ $\;\;$ [Given]
$AC$ is a transversal.
$\therefore \;$ $\angle DAF = \angle ECF$ $\;\;$ [Alternate angles] $\;\;\; \cdots \; (1)$
Also, $\;$ $\angle AFD = \angle EFC$ $\;\;$ [Vertically opposite angles] $\;\;\; \cdots \; (2)$
$\therefore \;$ From $(1)$ and $(2)$, $\;$ $\triangle ADF \sim \triangle CEF$ $\;\;$ [By Angle-Angle postulate] - $\because \;$ $\triangle ADF \sim \triangle CEF$,
$\dfrac{AD}{CE} = \dfrac{AF}{CF}$ $\;\;$ [corresponding sides of similar triangles are proportional]
$\implies$ $AD = CE \times \dfrac{AF}{CF}$ $\;\;\; \cdots \; (3)$
Given: $\;$ $\dfrac{AF}{AC} = \dfrac{5}{8}$
$\implies$ $1 - \dfrac{AF}{AC} = 1 - \dfrac{5}{8}$
i.e. $\;$ $\dfrac{AC - AF}{AC} = \dfrac{CF}{AC} = \dfrac{3}{8}$
i.e. $\;$ $\dfrac{AC}{CF} = \dfrac{8}{3}$
$\therefore \;$ $\dfrac{AF}{AC} \times \dfrac{AC}{CF} = \dfrac{5}{8} \times \dfrac{8}{3}$
i.e. $\;$ $\dfrac{AF}{CF} = \dfrac{5}{3}$ $\;\;\; \cdots \; (4)$
Given: $\;$ $CE = 6 \; cm$ $\;\;\; \cdots \; (5)$
In view of equations $(4)$ and $(5)$, equation $(3)$ becomes
$AD = 6 \times \dfrac{5}{3} = 10 \; cm$ - Given: $\;$ $DE \parallel BC$ $\implies$ $DF \parallel BC$
$AB$ is a transversal.
$\therefore \;$ $\angle ADF = \angle ABC$ $\;\;$ [Corresponding angles] $\;\;\; \cdots \; (6)$
$\angle BAC = \angle DAF$ $\;\;$ [Common angle] $\;\;\; \cdots \; (7)$
$\therefore \;$ From $(6)$ and $(7)$, $\;$ $\triangle ADF \sim \triangle ABC$ $\;\;$ [By Angle-Angle postulate]
$\therefore \;$ $\dfrac{\text{Area } \left(\triangle ADF\right)}{\text{Area } \left(\triangle ABC\right)} = \dfrac{AF^2}{AC^2}$
[Areas of two similar triangles are proportional to the squares of their corresponding sides]
$\therefore \;$ $\dfrac{\text{Area } \left(\triangle ADF\right)}{\text{Area } \left(\triangle ABC\right)} = \left(\dfrac{5}{8}\right)^2 = \dfrac{25}{64}$