Parallels, Midpoints and the Bisector Theorem

If $D$ is on $AB$, $E$ on $AC$ and $DE \parallel BC$:

• $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ (the basic proportionality theorem);
• $\dfrac{AD}{AB} = \dfrac{AE}{AC} = \dfrac{DE}{BC}$ (the small triangle $ADE$ is similar to $ABC$).

Converse: if a line divides two sides in the same ratio, it is parallel to the third side.

• The segment joining the midpoints of two sides is parallel to the third side and half as long.
• Converse: a line through the midpoint of one side, parallel to another side, bisects the third side.
• Joining the three midpoints makes four congruent triangles, each with half the perimeter of the original.
• Lines through each vertex parallel to the opposite side make a triangle with the original vertices as its midpoints, so its sides are double.
• If a triangle is cut into three corner triangles and one middle triangle, the corner perimeters add up to the outer perimeter plus the middle perimeter.

If $AD$ bisects $\angle A$ and meets $BC$ at $D$:

• $\dfrac{BD}{DC} = \dfrac{AB}{AC}$, so $BD = \dfrac{a\,c}{b + c}$ and $DC = \dfrac{a\,b}{b + c}$ (with $a = BC$, $b = CA$, $c = AB$);
• the areas of $ABD$ and $ACD$ are in the ratio $AB : AC$.

Converse: if $AB\cdot DC = AC\cdot BD$, then $AD$ bisects $\angle A$.