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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.

Suppose $CM$ is the perpendicular bisector of $AB,$ and that $M$ is the midpoint of $AB.$

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, $CM.$

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 |

$ AM≅MBAB⊥CM $ | Given |

$△AMC≅△BMC$ | SAS congruence theorem |

$AC≅BC$ | $△AMC≅△BMC$ |

$AC=BC$ | Definition of congruent segments |

Determine the perimeter of $△ABC.$

Show Solution

To determine the perimeter, we need the lengths of each side of the triangle. The line $CD$ divides $AB$ in two equal parts at a right angle, meaning it's a perpendicular bisector. According to the Perpendicular Bisector Theorem, the segments $AC$ and $BC$ are congruent. Therefore,
$2x+1=13−x.$
Let's solve for $x.$
Since $x=4,$ $AD$ is $4$ units long. The segment $BD$ is congruent to $AD,$ which means that $AB$ in total measures $8$ units. We can use $x=4$ to find the lengths of $AC.$
$AC=2x+1⇔AC=2⋅4+1⇔AC=9$
Since $AC≅BC,$ $BC$ 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.

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 $DE$ is a midsegment of $△ABC,$ $DE∥AB$ and $DE=21 AB.$

If a segment parallel to one of the sides of a triangle is drawn between the other sides, the segment divides the other two sides proportionally.

Based on the diagram, the following relation holds true.

If $DE∥AB,$ then $DCAD =ECBE $

Since $DE$ and $AB$ are parallel, by the Corresponding Angles Theorem, $∠CDE$ and $∠CAB$ are congruent. Similarly, $∠CED$ and $∠CBA$ are congruent.

Therefore, by the Angle-Angle Similarity Theorem, $△ABC$ and $△DEC$ are similar. Consequently, their corresponding sides are proportional. $△ABC∼△DEC⇓DCAC =ECBC $ Applying the Segment Addition Postulate, both numerators can be rewritten. $ACBC =AD+DC=BE+EC $ Substituting these expressions into the equation above, the required proportion will be obtained.$DCAC =ECBC $

$DCAD+DC =ECBE+EC $

WriteSumFracWrite as a sum of fractions

$DCAD +DCDC =ECBE +ECEC $

SimpQuotSimplify quotient

$DCAD +1=ECBE +1$

SubEqn$LHS−1=RHS−1$

$DCAD =ECBE $

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.

Consider $△ABC.$ The points $D$ and $E$ are midpoints on their respective side. Thus, $AE$ and $BD$ are medians.

The two medians intersect at the point $F.$

Now, two new points are introduced — the midpoints of $AF$ and $BF.$ Call them $G$ and $H.$

Since $G$ and $H$ are midpoints of $AF$ and $BF,$ $GH$ is a midsegment of $△AFB.$ Thus, by the Triangle Midsegment Theorem, $GH$ is parallel to and half the length of $AB.$

Similarly, $DE$ is a midsegment of $△ABC$ since $D$ and $E$ are the midpoints of $AC$ and $BC.$ Therefore, $DE$ is also parallel to and half the length of $AB.$ It follows that $GH∥DEandGH=DE.$

Since $DE$ and $GH$ 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, $DF≅FH$ and $GF≅FE.$

Thus, the median $DB$ intersects $AE$ at two-thirds of the distance from $A.$ Now, by applying the same reasoning for the third median, it also intersects $AE$ at two-thirds from $A.$

The median from point $C$ also intersects $AE$ 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 $AD.$

Show 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 $AD.$
Now we can use the value of $x$ to find the length of $AD.$
$ADADADAD =(8−2x)+x=8−x=8−2=6 $
The length of $AD$ is $6$ units.

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