Question

Find A($△$DOE) : A($△$DCE)

Answer 1

In $△$s ADE and ABC,
$\angle DAE=\angle BAC$ (Common angle)
$\angle ADE=\angle ABC$ (Corresponding angles of parallel lines)
$\angle AED=\angle ACB$ (Corresponding angles of parallel lines)
Therefore, $△ADE\sim △ABC$ (AAA rule)
Hence, $\frac{AD}{AB}=\frac{DE}{BC}$
$\frac{AD}{AD+DB}=\frac{DE}{BC}$
$\frac{1}{1+\frac{DB}{AD}}=\frac{DE}{BC}$
$\frac{1}{1+\frac{4}{5}}=\frac{DE}{BC}$
$\frac{5}{9}=\frac{DE}{BC}$
Now, In $△$s, ODE and OBC,
$\angle DOE=\angle BOC$ (Vertically opposite angles)
$\angle DEO=\angle OBC$ (Alternate angles of parallel lines)
$\angle EDO=\angle OCB$ (Alternate angles of parallel lines)
Hence, $△ODE\sim △OCB$ (AAA rule)
$\therefore$ $\frac{OC}{OD}=\frac{BC}{DE}$
$\frac{OC}{OD}+1=\frac{9}{5}+1$
$\frac{OC+OD}{OD}=\frac{9+5}{5}$
$\frac{DC}{OD}=\frac{14}{5}$
$\frac{OD}{DC}=\frac{5}{14}$

Now, from E drop a perpendicular on CD such that EN $\perp$ CD
$\frac{Ar.△DOE}{Ar.△DCE}=\frac{\frac{1}{2}DO\times EN}{\frac{1}{2}DC\times EN}$
$\frac{Ar.△DOE}{Ar.△DCE}=\frac{DO}{DC}=\frac{5}{14}$