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{{ courseTrack.displayTitle }} {{ printedBook.courseTrack.name }} {{ printedBook.name }} # Dissecting Triangles

Every triangle possesses characteristics that can be used to further analyze the triangle. These include medians, midsegments, and the centroid.
Rule

## Perpendicular Bisector Theorem

Any point on a perpendicular bisector is equidistant from the endpoints of the line segment. This can be proven using congruent triangles.

### Proof

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Perpendicular Bisector Theorem

Suppose $\overline{CM}$ is the perpendicular bisector of $\overline{AB},$ and that $M$ is the midpoint of $\overline{AB}.$ Two triangles can be created by connecting points $A$ and $C,$ and $B$ and $C.$ These triangles both have a right angle and one of the legs measures half of $AB.$ They also share one leg, $\overline{CM}.$ According to the SAS Congruence Theorem, the triangles are congruent. Thus, their hypotenuses are also congruent. Therefore, any point on a perpendicular bisector is equidistant from the endpoints of the segment. This can be summarized in a two-column proof.

 Statement Reason \begin{aligned}&\overline{AM} \cong \overline{MB} \\ &\overline{AB} \perp \overleftrightarrow{CM} \end{aligned} Given $\triangle AMC \cong \triangle BMC$ SAS congruence theorem $\overline{AC} \cong \overline{BC}$ $\triangle AMC \cong \triangle BMC$ $AC = BC$ Definition of congruent segments
Note that $AMC$ and $MBC$ are not triangles if $C$ is the point of intersection, $M.$ However, since $M$ is the midpoint of $\overline{AB}$ it is, by definition, equidistant from $A$ and $B.$
Exercise

Determine the perimeter of $\triangle ABC.$ Solution
To determine the perimeter, we need the lengths of each side of the triangle. The line $\overleftrightarrow{CD}$ divides $\overline{AB}$ in two equal parts at a right angle, meaning it's a perpendicular bisector. According to the Perpendicular Bisector Theorem, the segments $\overline{AC}$ and $\overline{BC}$ are congruent. Therefore, $2x+1=13-x.$ Let's solve for $x.$
$2x+1=13-x$
$3x+1=13$
$3x=12$
$x=4$
Since $x=4,$ $AD$ is $4$ units long. The segment $\overline{BD}$ is congruent to $\overline{AD},$ which means that $AB$ in total measures $8$ units. We can use $x=4$ to find the lengths of $\overline{AC}.$ $AC=2x+1 \Leftrightarrow AC=2 \cdot {\color{#0000FF}{4}} +1 \Leftrightarrow AC=9$ Since $\overline{AC} \cong \overline{BC},$ $\overline{BC}$ also measures $9$ units. We can now add all three side lengths to determine the perimeter. $AB +BC + AC = 8+9+9=26$ The perimeter of the triangle is $26$ units.
info Show solution Show solution
Concept

## Midsegment

A midsegment is a segment that connects the midpoints of two of the sides of a triangle. Since a triangle has three sides, there are three midsegments. Rule

## Triangle Midsegment Theorem

The segment that connects the midpoints of two sides of a triangle — a midsegment — is parallel to the third side of the triangle and half the length. If $\overline{DE}$ is a midsegment of $\triangle ABC,$ $\overline{DE} \parallel \overline{AB}$ and $DE=\frac{1}{2}AB.$

This can be proven in the coordinate plane.

### Proof

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Triangle Midsegment Theorem

If $\overline{DE}$ is a midsegment of $\triangle ABC,$ $\overline{DE} \parallel \overline{AB}$ and $DE=\frac{1}{2}AB.$

Proof

Draw an arbitrary triangle, $\triangle ABC$ in the coordinate plane so that $A$ lies on the origin, and $\overline{AB}$ is horizontal and lies on the $x$-axis. Since $A$ lies on the origin, its coordinates are $(0,0).$ Point $B$ is on the $x$-axis, meaning its $y$-coordinate is $0.$ The remaining coordinates are unknown, and can be named $a,$ $b,$ and $c.$ Then, the points are $A(0,0), \quad B(a,0), \quad \text{and} \quad C(b,c).$ Draw midsegment $\overline{DE}$ from $\overline{AC}$ to $\overline{BC}.$ By defintion, $D$ is the midpoint of $\overline{AC}$ and $E$ the midpoint of $\overline{BC}.$ Proof

### $\overline{DE}$ parallel to $\overline{AB}$

If the slopes of $\overline{DE}$ and $\overline{AB}$ are equal, the segments are parallel. Since $\overline{AB}$ was drawn as a horizontal line, its slope equals $0.$ To show that the slope of $\overline{DE}$ also equals $0,$ it can be shown that the $y$-coordinates of its points are the same. To do this, $A$ and $C$ and the midpoint formula can be used. \begin{aligned} y\text{-coordinate of }D: y_D=\frac{y_1+y_2}{2} \Leftrightarrow y_D=\frac{0+c}{2} \Leftrightarrow y_D=\frac{c}{2} \\ y\text{-coordinate of }E: y_E=\frac{y_1+y_2}{2} \Leftrightarrow y_E=\frac{0+c}{2} \Leftrightarrow y_E=\frac{c}{2} \end{aligned} The $y$-coordinates of $D$ and $E$ are both $\frac{c}{2}.$ It follows that the slope of $\overline{DE}$ equals $0.$ Thus, since $\overline{AB}$ and $\overline{DE}$ have equal slopes, they are parallel.

Proof

### $\overline{DE}$ is half the length of $\overline{AB}$

Since both $\overline{AB}$ and $\overline{DE}$ are horizontal, their lengths are given by the difference of the endpoints' $x$-values. The $x$-coordinates of $A$ and $B$ are $0$ and $a,$ respectively. This gives $AB = a-0=a.$ To find the $x$-coordinates of $D$ and $E,$ the midpoint formula is used. \begin{aligned} x\text{-coordinate of }D &: x_D=\frac{x_1+x_2}{2} \Leftrightarrow x_D=\frac{0+b}{2} \Leftrightarrow x_D=\frac{b}{2} \\ x\text{-coordinate of }E &: x_E=\frac{x_1+x_2}{2} \Leftrightarrow x_E=\frac{b+a}{2} \Leftrightarrow x_E=\frac{b}{2}+\frac{a}{2} \end{aligned} The length of $\overline{DE}$ is given by subtracting the $x$-coordinates: $DE=\dfrac{b}{2}+\dfrac{a}{2}-\dfrac{b}{2}=\dfrac{a}{2}.$ Since $DE=\frac{a}{2}$ is half of $AB=a,$ the midsegment $\overline{DE}$ is half the length of $\overline{AB}.$

Therefore, a midsegment of a triangle is parallel to the third side of the triangle and half the length.
Rule

## Triangle Proportionality Theorem

If a segment parallel to one of the sides in a triangle is drawn between the other sides, the segment divides the other two sides proportionally. For $\triangle ABC,$ since $\overline{DE} \parallel \overline{AB},$ $\dfrac{AD}{DC}=\dfrac{BE}{EC}.$

This can be proven using congruent angles.

### Proof

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Triangle Proportionality Theorem

Since $\overline{DE}$ and $\overline{AB}$ are parallel, the corresponding angles $\angle CDE$ and $\angle CAB$ are congruent. Similarly, $\angle CED$ and $\angle CBA$ are congruent. Also, triangles $\triangle ABC$ and $\triangle CDE$ share the angle $\angle C.$ This means that $\triangle ABC$ and $\triangle CDE$ are the same shape, but $\triangle CDE$ is a smaller version. Thus, the ratios between corresponding sides of the triangles are equal. This leads to $\dfrac{AC}{DC}=\dfrac{BC}{EC}.$ Using that $AC=AD+DC$ and $BC=BE+EC,$ it's possible to rearrange this equality to get the proportionality theorem.
$\dfrac{AC}{DC}=\dfrac{BC}{EC}$
$\dfrac{{\color{#0000FF}{AD+DC}}}{DC}=\dfrac{{\color{#009600}{BE+EC}}}{EC}$
$\dfrac{AD}{DC}+\dfrac{DC}{DC}=\dfrac{BE}{EC}+\dfrac{EC}{EC}$
$\dfrac{AD}{DC}+1=\dfrac{BE}{EC}+1$
$\dfrac{AD}{DC}=\dfrac{BE}{EC}$
Thus, the following equality yields $\dfrac{AD}{DC}=\dfrac{BE}{EC}.$ Therefore, a segment parallel to one side of a triangle divides the other two sides proportionally. This can be summarized in a flowchart proof. Concept

## Median

A median of a triangle is a line segment between the midpoint of one side and the opposite vertex. Since a triangle has three vertices, every triangle has three medians. All medians intersect each other at one point, the centroid.
Concept

## Centroid

The three medians of a triangle intersect at one point called the centroid. Rule

## Centroid Theorem

The centroid of a triangle is two-thirds of the distance from each vertex to the midpoint of the opposite side. If $\overline{AE},$ $\overline{BD},$ and $\overline{CF}$ are the medians of $\triangle ABC,$ then $AP=\tfrac{2}{3}AE, BP=\tfrac{2}{3}BD, \text{ and } CP=\tfrac{2}{3}CF.$

This can be proven using midpoints and parallel lines.

### Proof

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Centroid Theorem

Consider $\triangle ABC.$ The points $D$ and $E$ are midpoints on their respective side. Thus, $\overline{AE}$ and $\overline{BD}$ are medians. The two medians intersect at the point $F.$ Now, two new points are introduced — the midpoints of $\overline{AF}$ and $\overline{BF}.$ Call them $G$ and $H.$ Since $G$ and $H$ are midpoints of $\overline{AF}$ and $\overline{BF},$ $\overline{GH}$ is a midsegment of $\triangle AFB.$ Thus, by the Triangle Midsegment Theorem, $\overline{GH}$ is parallel to and half the length of $\overline{AB}.$ Similarly, $\overline{DE}$ is a midsegment of $\triangle ABC$ since $D$ and $E$ are the midpoints of $\overline{AC}$ and $\overline{BC}.$ Therefore, $\overline{DE}$ is also parallel to and half the length of $\overline{AB}.$ It follows that $\overline{GH} \parallel \overline{DE} \quad \text{and} \quad GH=DE.$ Since $\overline{DE}$ and $\overline{GH}$ are parallel and congruent, $DEHG$ are the vertices of a parallelogram. By the Parallelogram Diagonals Theorem, the diagonals of a parallelogram bisect each other. Therefore, $\overline{DF}\cong \overline{FH}$ and $\overline{GF}\cong \overline{FE}.$ Thus, the median $\overline{DB}$ intersects $\overline{AE}$ at two-thirds of the distance from $A.$ Now, by applying the same reasoning for the third median, it also intersects $\overline{AE}$ at two-thirds from $A.$ The median from point $C$ also intersects $\overline{AE}$ at two-thirds from point $A.$ Therefore, the centroid of a triangle is two-thirds the distance from each vertex to the midpoint of the opposite side. Exercise

Determine the length of $\overline{AD}.$ Solution
To begin, notice that the segments inside the triangle each bisect their respective sides. Thus, they are medians and intersect at the centroid. By the Centroid Theorem, the sides that measures $8-2x$ is twice as long as the segment that measures $x.$ We can write the following equation. $8-2x=2x.$ Solving for $x$ will allow us to determine the length of the smaller segments, which we can then use to find the length of $\overline{AD}.$
$8-2x=2x$
$8=4x$
$4x=8$
$x=2$
Now we can use the value of $x$ to find the length of $\overline{AD}.$ \begin{aligned} AD&=(8-2x)+x\\ AD&=8-x\\ AD&=8-{\color{#0000FF}{2}}\\ AD&=6 \end{aligned} The length of $\overline{AD}$ is $6$ units.
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