Question

In the given figure, $FG//DE//BC.$ If $AF=6cm,FD=10cm,DB=4cm$ and $AE=12cm,$ find the length of $AG$

Answer 1

Given: $FG\parallel DE\parallel BC$

Now, In $△AFG$ and $△ADE$,
$\angle FAG=\angle DAE$ (Common angle)
$\angle AFG=\angle ADE$ (Corresponding angles)
$\angle AGF=\angle AED$ (Corresponding angles)
Thus, $△AFG\sim △ADE$ (AAA rule)

Similarly, $△AFG\sim △ABC$ and $△ADE\sim △ABC$

Now, since, $△AFG\sim △ADE$
$\frac{AF}{AD}=\frac{AG}{AE}$
$\frac{AF}{AF+FD}=\frac{AG}{AE}$
$\frac{6}{6+10}=\frac{AG}{12}$
$AG=\frac{6\times 12}{16}$
$AG=4.5$ cm

$AE=AG+GE$
$12=4.5+GE$
$GE=7.5$ cm

since, $△ADE\sim △ABC$
$\frac{AD}{AB}=\frac{AE}{AC}$
$\frac{AF+FD}{AF+FD+DB}=\frac{AE}{AE+EC}$
$\frac{6+10}{6+10+4}=\frac{12}{12+EC}$
$12+EC=\frac{12\times 20}{16}$
$12+EC=15$
$EC=3$ cm