Question

In the given figure, ABC and CEF are two triangles where BA is parallel to CE and AF : AC = 5 : 8. [3 MARKS]

i) Prove that $\Delta ADF\sim \Delta CEF$
ii) Find AD, if CE = 6 cm
iii) If DF is parallel to BC, find area of $\Delta ADF$ : area of $\Delta ABC$

Answer 1

i) In $\Delta ADF$ and $\Delta CFE$,
$\angle DAF=\angle ECF$ [Alternate angles]
$\angle AFD=\angle CFE$ [Vertically opposite angles]
$\therefore \angle ADF=\angle CEF$
$\Delta ADF\sim \Delta CEF$ [By AAA similarity]
Hence, proved.$\left[1Mark\right]$

ii) Since the corresponding sides of similar triangles are proportional.
$\therefore$ In $\Delta ADF$ and $\Delta CEF$,
$\frac{AD}{CE}=\frac{AF}{FC}\Rightarrow \frac{AD}{6}=\frac{5}{8-5}$
$\Rightarrow \frac{AD}{6}=\frac{5}{3}$
$\Rightarrow AD=\frac{30}{3}=10cm$$\left[1Mark\right]$

iii) If DF is parallel to BC, then $\Delta ADF\sim \Delta ABC$
Now, $\frac{ar\left(\Delta ADF\right)}{ar\left(\Delta ABC\right)}=\frac{A{F}^2}{A{C}^2}$
$=\frac{5^2}{8^2}=\frac{25}{64}$$\left[1Mark\right]$