Question

In given figure, CD and GH are respectively the medians of $△ABC$ and $△EFG$. If $△ABC\sim △FEG$, Prove that

(i) $△ADC\sim △FHG$

(ii) $\frac{CD}{GH}=\frac{AB}{FE}$

(iii) $△CDB\sim △GHE$

Answer 1

It is given that CD and GD are medians of ${△}^{\prime }sABC$ and $EFG$ respectively.

$\therefore$ $2AD=AB$ and $2FH=FE$......(i)

It is also given that $△ABC\sim △FEG$

$\therefore$ $\frac{AB}{FE}=\frac{AC}{FG}=\frac{BC}{EG}$ and, $\angle A=\angle F,\angle B=\angle E,\angle C=\angle G$...........(ii)

Now, $\frac{AB}{FE}=\frac{AC}{FG}=\frac{BC}{EG}$

$\Rightarrow$ $\frac{2AD}{2FH}=\frac{AC}{FG}=\frac{BC}{EG}$ [Using (i)]

$\Rightarrow$ $\frac{AD}{FH}=\frac{AC}{FG}=\frac{BC}{EG}$.........(iii)

(I) In ${△}^{\prime }sADC$ and $FHG$, we have

$\frac{AD}{FH}=\frac{AC}{FG}$ [From (iii)]

and , $\angle A=\angle F$

So, by SAS criterion of similarity, we have

$△ADC\sim △FHG$ $\left[Henceproved\right]$

(ii) we have,

$△ADC\sim △FHG$ [Proved above]

$\Rightarrow$ $\frac{DC}{HG}=\frac{AD}{FH}$

$\Rightarrow$ $\frac{CD}{GH}=\frac{2AD}{2FH}$

$\Rightarrow$ $\frac{CD}{GH}=\frac{AB}{FE}$ [$\because$ AB=2AD and FE=2FH] $\left[Henceproved\right]$

(iii) We have,

$\frac{AB}{FE}=\frac{AC}{FG}=\frac{BC}{EG}$ [From (i)]

Also, $\frac{CD}{GH}=\frac{AB}{FE}$ [As proved above]

$\therefore$ $\frac{CD}{GH}=\frac{BC}{EG}$.......(iv)

Again, $\frac{AB}{FE}=\frac{AC}{FG}=\frac{BC}{EG}$

$\Rightarrow$ $\frac{2DB}{2HE}=\frac{BC}{EG}$ [$\because$ D and H are mid-points of AB and FE respectively]

$\Rightarrow$ $\frac{DB}{HE}=\frac{BC}{EG}$

From (iv) and (v), we have

$\frac{CD}{GH}=\frac{BC}{EG}=\frac{DB}{HE}$

$\Rightarrow$ $\frac{CD}{GH}=\frac{DB}{HE}=\frac{BC}{EG}$

$\Rightarrow$ $△CDB\sim △GHE$ [By SSS criterion of similarity] $\left[Henceproved\right]$