Question

Consider two triangular parks ABC and EFG such that a path from one corner of the park bisects the other side of the park as shown in the figure. If the dimensions of the park ABC are AB = 4 m, BC = 6 m and AC = 3 m and that of park EFG are EF = 2 m, EG = 3 m and FG = 1.5 m.

Then, which of the following is true?

• A. $\Delta ADC\sim \Delta FHG$
  
• B. $\frac{CD}{GH}=\frac{AB}{FE}$
• C. $\Delta CDB\sim GHE$
• D. All of the above

Answer 1

The correct option is D

All of the above

In $\Delta ABC$ and $\Delta EFG$
AB = 4 m
BC = 6 m
AC = 3 m
EF = 2 m
FG = 1.5 m
and EG = 3 m

CD and GH are medians to sides AB and EF, respectively.
Now in $\Delta ABC$ and $\Delta EFG$,
$\frac{AB}{BC}=\frac{EF}{EG}$
and $\frac{BC}{AC}=\frac{EG}{FG}$
$\therefore \Delta ABC\sim \Delta FEG\dots \left(i\right)$ [by SSS similarity]

$\Rightarrow \angle A=\angle F,\angle B=\angle E\&\angle C=\angle G\dots \left(ii\right)$

Now, $\frac{AB}{FE}=\frac{AC}{FG}=\frac{BC}{EG}\Rightarrow \frac{2AD}{2FH}=\frac{AC}{FG}=\frac{BC}{EG}\Rightarrow \frac{AD}{FH}=\frac{AC}{FG}=\frac{BC}{EG}\dots \left(iii\right)$

(a) In $\Delta ADCand\Delta FHG$, we have
$\frac{AD}{FH}=\frac{AC}{FG}$
and $\angle A=\angle F$ (from Eq. (iii))
$\therefore \Delta ADC\sim \Delta FHG$ [by SAS similarity]

(b) $\text{We have,}\Delta ADC\sim \Delta FHG\Rightarrow \frac{DC}{HG}=\frac{AD}{FH}\Rightarrow \frac{CD}{GH}=\frac{2AD}{2FH}\Rightarrow \frac{CD}{GH}=\frac{AB}{FE}[\because AB=2AD,FE=2FH]$

(c) Now, $\frac{AB}{FE}=\frac{AC}{FG}=\frac{BC}{EG}\dots \left(iv\right)$ [from Eq. (i)]
Also, $\frac{CD}{GH}=\frac{AB}{FE}$
$\Rightarrow \frac{CD}{GH}=\frac{BC}{EG}$ [from Eq. (iv)]
$\therefore \Delta CDB\sim \Delta GHE$ [by SSS similarity]