Bisectors and Similarities in Triangles

SAT Questions that focus on Bisectors and Similarities in Triangles require knowledge of the following topics.

Parallel lines cut by two transversals

Consider the parallel lines (s1, s2...) and two transversals (t1 e t2) (figure below):

Theorem: The parallel lines divide the transversals proportionally, that is:

$\frac{CD}{C'D'} = \frac{BC}{B'C'} = \frac{AB}{A'B'}$

In order to prove this theorem, let's divide segments CD using "n" parallel lines, and segment BC using "p" parallel lines, so that the distance between all of these parallel lines is constant.

Thus, segment CD will be divided in "n" segments with size x; C'D', in "n" segments with size x'. Segment BC will be divided in "p" segments with size x; B'C', in "p" segments with size x'.

So we have that:

$\frac{CD}{C'D'} = \frac{BC}{B'C'}$

$\frac{nx}{nx'} = \frac{px}{px'}$

$\frac{x}{x'} = \frac{x}{x'}$, which proves the theorem.

Interior Angle Bisector Theorem

Consider triangle ABC (figure below), in which the red line is the bisector that divides angle  into two equal angles "y".

The dotted line DB is parallel to the bisector of angle Â. Side AB and segment DC are transversals of these two parallel lines.

In triangle ADB, angles D and B are equal to "y" (angles of a transversal). Given that these two angles are congruent, triangle ADB is isosceles: AD = AB = "c".

Using the theorem of parallel lines cut by two transversals:

$\frac{AD}{m}=\frac{AC}{n}$

$\frac{c}{m}=\frac{b}{n}$

$\frac{c}{b}=\frac{m}{n}$

Exterior Angle Bisector Theorem

Consider triangle ABC in the following figure, in which the red line is the bisector that divides the exterior angle  into two equal angles "y".

The dotted line EB is parallel to the bisector of exterior angle Â. Thus side AC and segment FC are transversals of these two parallel lines.

In triangle ABE, angles B and E are equal to "y" (angles of a transversal). Given that these two angles are congruent, triangle ABE is isosceles: AE = AB = "c".

Using the theorem of parallel lines cut by two transversals:

$\frac{AC}{n+a}=\frac{AE}{n}$

$\frac{b}{m}=\frac{c}{n}$

$\frac{c}{b}=\frac{m}{n}$, where "m" is equal to $FC = n+a$.

90º Angle between the Interior and the Exterior Angle Bisectors

Consider the following figure, in which the red lines are the bisectors of the interior and of the exterior angles in vertex C:

The interior angle is divided in two equal angles "r" and "r"; the exterior angle is divided in two equal angles "s" and "s".

The sum of the interior and exterior angles in any vertex is 180º. Thus:
$r+r+s+s=180º$
$2r+2s=180º$
$r+s=90º$

Parallel Lines and Triangles

Consider triangles ABC and AB'C' in the following figure, where sides B'C' and BC are parallel:

Sides AB and AC are two transversals that cut the two parallel sides (B'C' and BC). Therefore:

$\frac{AB'}{AC'}=\frac{AB}{AC}$

$\frac{AB'}{AB}=\frac{AC'}{AC}$

Solved SAT Practice Tests

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SAT Practice Tests - Bisectors and Similarities in Triangles and

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