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

Any point on a perpendicular bisector is equidistant from the endpoints of the line segment.

This can be proven using congruent triangles. ## Proof | info | |

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 |

Determine the perimeter of $\triangle ABC.$

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 $x.$
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.

$2x+1=13-x$

$3x+1=13$

$3x=12$

$x=4$

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.

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 $DE=\frac{1}{2}AB.$

## Proof | info | |

Triangle Midsegment Theorem |

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

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 $\overline{BC}.$

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.

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.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 | info | |

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.$\dfrac{AC}{DC}=\dfrac{BC}{EC}$

$\dfrac{{\color{#0000FF}{AD+DC}}}{DC}=\dfrac{{\color{#009600}{BE+EC}}}{EC}$

$\dfrac{AD}{DC}+\dfrac{DC}{DC}=\dfrac{BE}{EC}+\dfrac{EC}{EC}$

$\dfrac{AD}{DC}+1=\dfrac{BE}{EC}+1$

$\dfrac{AD}{DC}=\dfrac{BE}{EC}$

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

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 | info | |

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.

Determine the length of $\overline{AD}.$

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 $\overline{AD}.$
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.

$8-2x=2x$

$8=4x$

$4x=8$

$x=2$

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