Question

Given : AB // DE and BC //EF. Prove that :
(i) $\frac{AD}{DG}=\frac{CF}{FG}$
(ii) $\Delta$ DFG~ $\Delta$ ACG.

Answer 1

In $△$ABG,

we have $DE\parallel AB\Rightarrow \frac{GE}{EB}=\frac{DG}{AD}...\left(1\right)$

In $△$CBG, we have

$EF\parallel BC\Rightarrow \frac{GE}{EB}=\frac{FG}{CF}...\left(2\right)$

From (1) and (2)

$\frac{DG}{AD}=\frac{FG}{CF}\Rightarrow \frac{AD}{DG}=\frac{CF}{FG}$

Since $\frac{AD}{DG}=\frac{CF}{FG}$

Thus, DF divides AG and GC of $△$ACG in the same ratio.

therefore, by the converse of basic proportionality theorem, we have $DF\parallel AC$

In $△$DFG and $△$ACG

$\angle G=\angle G$ [Common]

$\angle GDF=\angle GAC$ [Corresponding angles]

$\angle GFD=\angle GCA$ [Corresponding angles]

Therefore, By AAA similarity $△DFG\sim △ACG$