Triangles - Basic Proportionality Theorem (Thales' Theorem)

The Basic Proportionality Theorem (BPT), also known as Thales' Theorem, states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio. In a triangle $\Delta ABC$, if a line $DE$ is drawn parallel to $BC$ intersecting $AB$ at $D$ and $AC$ at $E$, then $\frac{AD}{DB} = \frac{AE}{EC}$.

The Converse of the Basic Proportionality Theorem states that if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side. Visually, if you find a point $D$ on $AB$ and $E$ on $AC$ such that the ratio of segments $\frac{AD}{DB}$ equals $\frac{AE}{EC}$, you can conclude that $DE \parallel BC$.

Corollaries of BPT involve the relationship between the segments and the whole sides of the triangle. By adding $1$ to both sides of the BPT ratio, we can derive that $\frac{AB}{AD} = \frac{AC}{AE}$ and $\frac{AB}{DB} = \frac{AC}{EC}$. This is useful when the full length of the triangle sides is known instead of just the segments.

BPT applies to Trapeziums as well. Any line drawn parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally. If in a trapezium $ABCD$ where $AB \parallel DC$, a line $EF$ is drawn parallel to $AB$ (and thus $DC$) intersecting $AD$ at $E$ and $BC$ at $F$, then $\frac{AE}{ED} = \frac{BF}{FC}$.

Internal Bisector Theorem is a specialized application related to proportions. It states that the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. For $\Delta ABC$ with angle bisector $AD$ of $\angle A$, the ratio is $\frac{BD}{DC} = \frac{AB}{AC}$.

Similarity Connection: When $DE \parallel BC$ in $\Delta ABC$, the smaller triangle $\Delta ADE$ is similar to the larger triangle $\Delta ABC$. This means that not only are the sides divided proportionally, but the ratio of the parallel segments is also equal to the ratio of the sides: $\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$.