If a line intersects sides $AB$ and $AC$ of a $\Delta ABC$ at $D$ and $E$ respectively and is parallel to $BC$,prove that $\frac{AD}{AB} = \frac{AE}{AC}$ (see Figure).

(N/A) Given: In $\Delta ABC$,$DE || BC$,where $D$ and $E$ are points on $AB$ and $AC$ respectively.
According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,the other two sides are divided in the same ratio.
Therefore,$\frac{AD}{DB} = \frac{AE}{EC}$.
Taking the reciprocal of both sides:
$\frac{DB}{AD} = \frac{EC}{AE}$.
Adding $1$ to both sides:
$\frac{DB}{AD} + 1 = \frac{EC}{AE} + 1$.
Taking the common denominator:
$\frac{DB + AD}{AD} = \frac{EC + AE}{AE}$.
Since $DB + AD = AB$ and $EC + AE = AC$,we get:
$\frac{AB}{AD} = \frac{AC}{AE}$.
Taking the reciprocal again:
$\frac{AD}{AB} = \frac{AE}{AC}$.
Hence,it is proved.