Dissecting Triangles

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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

Perpendicular Bisector Theorem


Suppose CM\overline{CM} is the perpendicular bisector of AB,\overline{AB}, and that MM is the midpoint of AB.\overline{AB}.

Two triangles can be created by connecting points AA and C,C, and BB and C.C.

These triangles both have a right angle and one of the legs measures half of AB.AB. They also share one leg, CM.\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
AMMBABCM\begin{aligned}&\overline{AM} \cong \overline{MB} \\ &\overline{AB} \perp \overleftrightarrow{CM} \end{aligned} Given
AMCBMC\triangle AMC \cong \triangle BMC SAS congruence theorem
ACBC\overline{AC} \cong \overline{BC} AMCBMC\triangle AMC \cong \triangle BMC
AC=BCAC = BC Definition of congruent segments
Note that AMCAMC and MBCMBC are not triangles if CC is the point of intersection, M.M. However, since MM is the midpoint of AB\overline{AB} it is, by definition, equidistant from AA and B.B.
Exercise

Determine the perimeter of ABC.\triangle ABC.

Solution
To determine the perimeter, we need the lengths of each side of the triangle. The line CD\overleftrightarrow{CD} divides AB\overline{AB} in two equal parts at a right angle, meaning it's a perpendicular bisector. According to the Perpendicular Bisector Theorem, the segments AC\overline{AC} and BC\overline{BC} are congruent. Therefore, 2x+1=13x. 2x+1=13-x. Let's solve for x.x.
2x+1=13x2x+1=13-x
3x+1=133x+1=13
3x=123x=12
x=4x=4
Since x=4,x=4, ADAD is 44 units long. The segment BD\overline{BD} is congruent to AD,\overline{AD}, which means that ABAB in total measures 88 units. We can use x=4x=4 to find the lengths of AC.\overline{AC}. AC=2x+1AC=24+1AC=9 AC=2x+1 \Leftrightarrow AC=2 \cdot {\color{#0000FF}{4}} +1 \Leftrightarrow AC=9 Since ACBC,\overline{AC} \cong \overline{BC}, BC\overline{BC} also measures 99 units. We can now add all three side lengths to determine the perimeter. AB+BC+AC=8+9+9=26 AB +BC + AC = 8+9+9=26 The perimeter of the triangle is 2626 units.
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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 DE\overline{DE} is a midsegment of ABC,\triangle ABC, DEAB\overline{DE} \parallel \overline{AB} and DE=12AB.DE=\frac{1}{2}AB.

This can be proven in the coordinate plane.

Proof

Triangle Midsegment Theorem


If DE\overline{DE} is a midsegment of ABC,\triangle ABC, DEAB\overline{DE} \parallel \overline{AB} and DE=12AB.DE=\frac{1}{2}AB.

Proof

Draw an arbitrary triangle, ABC\triangle ABC in the coordinate plane so that AA lies on the origin, and AB\overline{AB} is horizontal and lies on the xx-axis.

Since AA lies on the origin, its coordinates are (0,0).(0,0). Point BB is on the xx-axis, meaning its yy-coordinate is 0.0. The remaining coordinates are unknown, and can be named a,a, b,b, and c.c. Then, the points are A(0,0),B(a,0),andC(b,c). A(0,0), \quad B(a,0), \quad \text{and} \quad C(b,c). Draw midsegment DE\overline{DE} from AC\overline{AC} to BC.\overline{BC}. By defintion, DD is the midpoint of AC\overline{AC} and EE the midpoint of BC.\overline{BC}.

Proof

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

If the slopes of DE\overline{DE} and AB\overline{AB} are equal, the segments are parallel. Since AB\overline{AB} was drawn as a horizontal line, its slope equals 0.0. To show that the slope of DE\overline{DE} also equals 0,0, it can be shown that the yy-coordinates of its points are the same. To do this, AA and CC and the midpoint formula can be used. y-coordinate of D:yD=y1+y22yD=0+c2yD=c2y-coordinate of E:yE=y1+y22yE=0+c2yE=c2\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 yy-coordinates of DD and EE are both c2.\frac{c}{2}. It follows that the slope of DE\overline{DE} equals 0.0. Thus, since AB\overline{AB} and DE\overline{DE} have equal slopes, they are parallel.

Proof

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

Since both AB\overline{AB} and DE\overline{DE} are horizontal, their lengths are given by the difference of the endpoints' xx-values. The xx-coordinates of AA and BB are 00 and a,a, respectively. This gives AB=a0=a. AB = a-0=a. To find the xx-coordinates of DD and E,E, the midpoint formula is used. x-coordinate of D:xD=x1+x22xD=0+b2xD=b2x-coordinate of E:xE=x1+x22xE=b+a2xE=b2+a2\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 DE\overline{DE} is given by subtracting the xx-coordinates: DE=b2+a2b2=a2. DE=\dfrac{b}{2}+\dfrac{a}{2}-\dfrac{b}{2}=\dfrac{a}{2}. Since DE=a2DE=\frac{a}{2} is half of AB=a,AB=a, the midsegment DE\overline{DE} is half the length of AB.\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 ABC,\triangle ABC, since DEAB,\overline{DE} \parallel \overline{AB}, ADDC=BEEC. \dfrac{AD}{DC}=\dfrac{BE}{EC}.

This can be proven using congruent angles.

Proof

Triangle Proportionality Theorem


Since DE\overline{DE} and AB\overline{AB} are parallel, the corresponding angles CDE\angle CDE and CAB\angle CAB are congruent. Similarly, CED\angle CED and CBA\angle CBA are congruent. Also, triangles ABC\triangle ABC and CDE\triangle CDE share the angle C.\angle C.

This means that ABC\triangle ABC and CDE\triangle CDE are the same shape, but CDE\triangle CDE is a smaller version. Thus, the ratios between corresponding sides of the triangles are equal. This leads to ACDC=BCEC. \dfrac{AC}{DC}=\dfrac{BC}{EC}. Using that AC=AD+DCAC=AD+DC and BC=BE+EC,BC=BE+EC, it's possible to rearrange this equality to get the proportionality theorem.
ACDC=BCEC\dfrac{AC}{DC}=\dfrac{BC}{EC}
AD+DCDC=BE+ECEC\dfrac{{\color{#0000FF}{AD+DC}}}{DC}=\dfrac{{\color{#009600}{BE+EC}}}{EC}
ADDC+DCDC=BEEC+ECEC\dfrac{AD}{DC}+\dfrac{DC}{DC}=\dfrac{BE}{EC}+\dfrac{EC}{EC}
ADDC+1=BEEC+1\dfrac{AD}{DC}+1=\dfrac{BE}{EC}+1
ADDC=BEEC\dfrac{AD}{DC}=\dfrac{BE}{EC}
Thus, the following equality yields ADDC=BEEC. \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 AE,\overline{AE}, BD,\overline{BD}, and CF\overline{CF} are the medians of ABC,\triangle ABC, then AP=23AE,BP=23BD, and CP=23CF. 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

Centroid Theorem


Consider ABC.\triangle ABC. The points DD and EE are midpoints on their respective side. Thus, AE\overline{AE} and BD\overline{BD} are medians.

The two medians intersect at the point F.F.

Now, two new points are introduced — the midpoints of AF\overline{AF} and BF.\overline{BF}. Call them GG and H.H.

Since GG and HH are midpoints of AF\overline{AF} and BF,\overline{BF}, GH\overline{GH} is a midsegment of AFB.\triangle AFB. Thus, by the Triangle Midsegment Theorem, GH\overline{GH} is parallel to and half the length of AB.\overline{AB}.

Similarly, DE\overline{DE} is a midsegment of ABC\triangle ABC since DD and EE are the midpoints of AC\overline{AC} and BC.\overline{BC}. Therefore, DE\overline{DE} is also parallel to and half the length of AB.\overline{AB}. It follows that GHDEandGH=DE. \overline{GH} \parallel \overline{DE} \quad \text{and} \quad GH=DE.

Since DE\overline{DE} and GH\overline{GH} are parallel and congruent, DEHGDEHG are the vertices of a parallelogram. By the Parallelogram Diagonals Theorem, the diagonals of a parallelogram bisect each other. Therefore, DFFH\overline{DF}\cong \overline{FH} and GFFE.\overline{GF}\cong \overline{FE}.

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

The median from point CC also intersects AE\overline{AE} at two-thirds from point A.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 AD.\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 82x8-2x is twice as long as the segment that measures x.x. We can write the following equation. 82x=2x. 8-2x=2x. Solving for xx will allow us to determine the length of the smaller segments, which we can then use to find the length of AD.\overline{AD}.
82x=2x8-2x=2x
8=4x8=4x
4x=84x=8
x=2x=2
Now we can use the value of xx to find the length of AD.\overline{AD}. AD=(82x)+xAD=8xAD=82AD=6\begin{aligned} AD&=(8-2x)+x\\ AD&=8-x\\ AD&=8-{\color{#0000FF}{2}}\\ AD&=6 \end{aligned} The length of AD\overline{AD} is 66 units.
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Exercises

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