Basic Proportionality Theorem

Statement:
If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.

To Prove:
\[\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\]

Proof:

1. A line parallel to a side of a triangle forms equal corresponding angles.

2. Hence, the two triangles formed are similar (AAA similarity).

3. In similar triangles, corresponding sides are proportional.

Therefore, the line divides the two sides in the same ratio.

\[\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\]

Statement:
If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

To Prove:
$DE\parallel BC$

$Proof:$

1. Assume a line through point D parallel to BC meets AC at F.

2. By BPT, \[\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AF}}{\mathrm{FC}}\]

3. Given, \[\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\]

4. Hence,\[\frac{AF}{FC}=\frac{AE}{EC}\]

5. ⇒ Points E and F coincide.

   Therefore,

   $DE\parallel BC$