12. More Theorems About Triangles
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Chapter 1
12.
More Theorems About Triangles
Discover more about triangles and their theorems. For example, the centroid is a unique point where all the medians of a triangle intersect, or the incenter, which is equidistant from each side of the triangle, making it the triangle's center. The orthocenter, is where the altitudes of a triangle meet. Depending on the type of triangle - acute, right, or obtuse - the orthocenter's position varies. These concepts not only enrich our understanding of triangles but also have practical applications. For instance, when designing a triangular table, placing the leg at the centroid ensures balance. Similarly, the incenter can guide the placement of items so they are equidistant from all sides.
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| | 12 Theory slides |
| | 9 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
More Theorems About Triangles
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The relationships within triangles can be investigated to understand the mystery behind the triangles. For this purpose, several theorems will be built upon the medians, bisectors, and altitudes of triangles will be investigated.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Investigating Medians of a Triangle
Using the following applet, draw the medians of △ ABC.
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Discussion
Centroid and Centroid Theorem
As it can be seen, medians of a triangle meet at a point. This point has a special name.
Concept
Centroid
The centroid of a triangle is the point of intersection of the triangle's medians. The centroid is typically represented by the letter G. This point is always inside the triangle.
By means of exploring the applet, it was also seen that the ratio ab is constant for each median.
Rule
Centroid Theorem
The centroid of a triangle is two-thirds the distance from each vertex to the midpoint of the opposite side.
If AE, BD, and CF are the medians of △ ABC, then the following statements hold true.
AG=2/3AE, BG=2/3BD, and CG=2/3CF
Proof
Consider a triangle with vertices A, B, and C as well as two of its medians. Let G be the point of intersection of the medians.
Let Q be a point on AC such that QE is parallel to BD.
In the diagram, ∠ EQC and ∠ BDC are corresponding angles. Since EQ and BD are parallel, these two angles are congruent by the Corresponding Angles Theorem. The same is true for ∠ QEC and ∠ DBC.
Therefore, △ ECQ and △ BCD have two pairs of congruent angles and are similar by the Angle-Angle Similarity Theorem. Similar reasoning can be used to show that △ AGD and △ AEQ are also similar. △ ECQ~△ BCD △ AGD~△ AEQ By the definition of a median, E is the midpoint of BC, and therefore, E divides BC into two congruent segments. Note that congruent segments have equal lengths. This information and the Segment Addition Postulate imply that the length of BC is two times the length of EC. BE=EC BC=BE+EC ⇒ BC=2EC Therefore, the scale factor of the similar triangles is 12. That means DC=2QC. Furthermore, by the Segment Addition Postulate, DC=DQ+QC. Then, using the Transitive Property of Equality and the Subtraction Property of Equality the following is obtained.
Since QC and DQ are equal, Q is the midpoint of DC.
Remembering that AD=DC, the ratio of DQ to AD can be calculated. DQ/AD=DQ/DC=1/2 Note that corresponding parts of similar triangles are proportional. Therefore, since △ AGD and △ AEQ are similar, the ratio of DQ to AD is equal to the ratio of GE to AG. GE/AG=1/2 ⇔ GE=1/2AG This information can be used to express AG in terms of AE.
AE=AG+GE
Substitute
GE= 1/2AG
AE=AG+ 1/2AG
▼
Solve for AG
NumberToFrac
a = 2* a/2
AE=2AG/2+1/2AG
MoveRightFacToNumOne
1/b* a = a/b
AE=2AG/2+AG/2
AddFrac
Add fractions
AE=3AG/2
MultEqn
LHS * 2=RHS* 2
2AE=3AG
DivEqn
.LHS /3.=.RHS /3.
2AE/3=AG
MovePartNumRight
a* b/c=a/c* b
2/3AE=AG
RearrangeEqn
Rearrange equation
AG=2/3AE ✓
This means that AG is two-thirds of AE. Now, consider △ ABC and its medians CF and AE. Let K be the point of intersection of these medians.
Let R be a point on AB such that ER is parallel to CF.
By following the same reasoning as before, it can be proved that AK is two-thirds of AE. Therefore, G and K are the same points. That means the medians are concurrent — they meet at one point.
Before it was shown that AG= 23AE. By using similar arguments, it can be also shown that BG= 23BD and CG= 23CF.
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Example
Triangular Table
Zosia is planning to throw a party in her new house. She wants to design a triangular table with one leg for the snacks and drinks. This design choice will ensure that no one while moving around, would bump into a table leg.
But wait, there is a problem she has to solve. She has no idea where to place the leg so that the table will be perfectly balanced. Lend some math skills and help her find the point on the table where the table leg should be placed.
Answer
See solution.
Solution
Recall that the centroid of a triangle is also the center of mass of the triangle. Therefore, if Zosia places the leg at the centroid, the table will be perfectly balanced. Using a ruler, Zosia can begin finding the centroid by drawing the median of a table side.
Since the centroid of a triangle is the point of intersection of the medians, the centroid will be on this median.
The Centroid Theorem states that the centroid of a triangle is two-thirds of the distance from each vertex to the midpoint of the opposite side. Using this theorem, the distance between the centroid and the vertex along the segment can be found. Notice that the length of the median is six feet. What is two-thirds of six feet?
2/3* 6 = 4
The centroid is 4 feet away from the vertex.
Therefore, if Zosia places the leg at this point, the table will be perfectly balanced.
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Investigating Perpendicular Bisectors of a Triangle
Use the applet to draw the perpendicular bisectors of the sides of △ ABC.
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Discussion
Circumcenter and Circumcenter Theorem
The perpendicular bisectors of the sides of a triangle are concurrent, as shown in the preceding exploration. The point of concurrency of the perpendicular bisectors of a triangle is known by a unique name.
Concept
Circumcenter
The circumcenter of a triangle is the point of intersection of the triangle's perpendicular bisectors. Circumcenter of a triangle is denoted by the letter S. It can be inside, outside, or on a triangle's side, depending on the triangle type.
The investigation also indicated that the distances from the circumcenter to each vertex of the triangle are equal.
Rule
Circumcenter Theorem
The circumcenter of a triangle is equidistant to the vertices of the triangle.
Based on the characteristics of the diagram, the following relation holds true.
AS=BS=CS
Proof
Assume that ABC is a triangle and DS, ES, and FS are the perpendicular bisectors of the sides of this triangle.
Notice that S is a point on the perpendicular bisector of AB. Therefore, by the Perpendicular Bisector Theorem, S is equidistant from A and B.
Similarly, S is also a point on the perpendicular bisector of BC. Using the Perpendicular Bisector Theorem once again, it can be concluded that S is equidistant from B and C.
By the Transitive Property of Equality, AP is equal to CP. AS= BS BS=CS ⇒ AS=CS This proves that AS, BS, and CS are all equal.
Two-Column Proof
The proof can be summarized in the following two-column table.
Statements
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Reasons
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1. 1. & ABCis a triangle & DSis a perpendicular bisector ofAB & ESis a perpendicular bisector ofBC & FSis a perpendicular bisector ofAC
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1. 1. Given
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2. 2. & AS=BS & BS=CS
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2. 2. Perpendicular Bisector Theorem
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3. 3. AS=CS
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3. 3. Transitive Property of Equality
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Investigating Angle Bisectors of a Triangle
This time, draw the angle bisectors of the interior angles of △ ABC.
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Discussion
Incenter and Incenter Theorem
Through exploration of the applet, it has been shown that the angle bisectors of a triangle intersect at one point.
Concept
Incenter
The incenter of a triangle is the point of intersection of the triangle's angle bisectors. The incenter is typically represented by the letter I. This point is considered to be the center of the triangle. For every triangle, the incenter is always inside the triangle.
Rule
Incenter Theorem
The incenter of a triangle is the point which is equidistant from each of the triangle's sides. This point is considered to be the center of the triangle.
Based on the diagram, the following relation holds true.
DI=EI=FI
Proof
Consider a triangle and its incenter I.
Let DI, EI, and FI be the distances from I to the sides of the triangle. Recall that the distance from a point to a segment is perpendicular to the segment.
By the definition of an incenter, AI is the angle bisector of ∠ BAC. Since I lies on AI, it is equidistant from the angle's sides by the Angle Bisector Theorem. DI=EI Similarly, since I lies on BI, which is the bisector of ∠ ABC, it is also equidistant from this angle's sides. EI=FI By bringing together the above information, the following is obtained. DI=EI and EI=FI ⇕ DI=EI=FI This means that I is equidistant from each of the triangle's sides.
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Example
Setting a Table
Now that Zosia has perfectly balanced her triangular table using the centroid, she is ready to put some snacks on it. The snacks should be equidistant from each side of the table so that her friends can reach them easily. To top it all off, Zosia wants to place a candle to illuminate the whole table. ¡Qué genial!
Where should she place the candle and snacks?
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Hint
Note that the candle should be equidistant from each corner of the table.
Solution
Since the snacks should be equidistant from each side of the table, begin by recalling the Incenter Theorem.
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Incenter Theorem |
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The incenter of a triangle is the point which is equidistant from each of the triangle's sides. This point is considered to be the center of the triangle. |
By this theorem, it can be concluded that the snacks should be placed in the incenter of the table. On the other hand, the candle should be equidistant from each corner of the table to illuminate the whole table. Therefore, consider the Circumcenter Theorem.
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Circumcenter Theorem |
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The circumcenter of a triangle is the point which is equidistant from each of the triangle's vertices. |
Therefore, the candle illuminates the whole table if Zosia place it in the circumcenter of the table. Note that the centroid of the table does not satisfy either of these locations because it helps to determine the location of the center of mass as in the previous example.
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Investigating Altitudes of a Triangle
After understanding the characteristics of the bisectors of a triangle, the relationship between the altitudes of a triangle will be investigated. Using the applet, explore what properties are related to the altitude. Begin by drawing the altitudes of △ ABC.
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Discussion
Orthocenter and the Orthocenter Theorem
As could be found in the previous exploration, when the altitudes of a triangle are drawn, they intersect at one point.
Concept
Orthocenter
The orthocenter of a triangle is the point where a triangle's altitudes intersect. It is usually denoted by the letter H.
Rule
Orthocenter Theorem
The altitudes of a triangle are concurrent. That is, they intersect at a common point.
Proof
Assume that △ ABC is a triangle with altitudes AK, BL, and CM.
Start by drawing lines that pass through the vertices A, B, and C. The lines should also be parallel to the opposite sides BC, AC, and AB, respectively.
Notice that ABCD and AEBC are parallelograms.
By the Parallelogram Opposite Sides Theorem, BC is congruent to both AD and EA. Since congruent segments have equal lengths, BC is equal to AD and EA. lBC≅AD BC≅EA ⇔ l BC = AD BC = EA By the Transitive Property of Equality, AD is equal to EA. BC=AD BC=EA ⇒ AD=EA Considering the analysis so far, it can be concluded that A is the midpoint of DE.
Now, consider the altitude AK from vertex A to the opposite side BC.
Since AK is perpendicular to BC and BC is parallel to ED, it can be concluded that AK is perpendicular to ED by the Perpendicular Transversal Theorem.
AK is perpendicular to ED and passes through its midpoint. By the definition of a perpendicular bisector, AK is the perpendicular bisector of ED.
Using this reasoning, it can be proved that the other altitudes of △ ABC, BL and CM, are perpendicular bisectors of EF and DF, respectively.
By the definition of circumcenter, the perpendicular bisectors AK, BL, and CM intersect at a point H. This implies that the altitudes of △ ABC are concurrent.
Two-Column Proof
The proof can be summarized in the following two-column table.
Statements
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Reasons
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1. 1. ABC is a triangle
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1. 1. Given
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2. 2. & DE∥ BC & EF∥ AC & DF∥ AB
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2. 2. Given (Drawn)
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3. 3. ABCD, AEBC, and ABFC are parallelograms
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3. 3. Definition of a parallelogram
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4. 4. & BC≅ AD, BC≅ EA & AC≅ BE, AC≅ FB & AB≅ DC, AB≅ CF
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4. 4. Parallelogram Opposite Sides Theorem
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5. 5. & BC=AD, BC=EA & AC=BE, AC=FB & AB=CF, AB=DC
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5. 5. Definition of congruent segments
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6. 6. & AD=EA & BE=FB & CF=DC
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6. 6. Transitive Property of Equality
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7. 7. & Ais the midpoint ofDE & Bis the midpoint ofEF & Cis the midpoint ofDF
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7. 7. Definition of a midpoint
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8. 8. AK, BL, and CM are the altitudes of △ ABC
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8. 8. Given
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9. 9. & AK⊥ BC & BL⊥ AC & CM⊥ AB
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9. 9. Definition of an altitude
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10. 10. & AK⊥ DE & BL⊥ EF & CM⊥ DF
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10. 10. Perpendicular Transversal Theorem
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11. 11. & AKis a perpendicular bisector ofDE & BLis a perpendicular bisector ofEF & CMis a perpendicular bisector ofDF
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11. 11. Definition of a perpendicular bisector
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12. 12. AK, BL, and CM are concurrent
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12. 12. Definition of circumcenter
|
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Closure
Euler's Line
In this lesson, relationships within triangles were covered. Four points have been introduced, along with their theorems. Did you know that there is a mysterious relationship between three of these points?
In any triangle, the centroid, circumcenter, and orthocenter of the triangle are collinear.
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More Theorems About Triangles
Exercise 2.1
In the following triangle, D is the centroid. Find the value of x if BD=4x+6 and BF=9x.
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To solve this exercise, recall the Centroid Theorem. In a triangle, the centroid is always two-thirds the distance from each vertex to the midpoint of the opposite side. Let's show this in the diagram.
We know that BD= 4x+6 and also that BF= 9x. Using this information and the Centroid Theorem, we can write an equation in terms of x. BD=2/3 BF ⇒ 4x+6=2/3( 9x) Let's solve this equation for x.
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Which of the following line(s) coincides with the median of an equilateral triangle ABC? A& The angle bisector B& The perpendicular bisector C& The altitude D& None of the above
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Let's draw an equilateral triangle. This is a triangle with three congruent sides and three congruent angles.
Drawing the Median
To draw a median of a triangle, we connect a vertex with the midpoint on the opposite side. Notice that since the median intersects the midpoint, it must divide the side into two equal halves.
Analyzing △ ABD and △ CBD
Examining the diagram, we see that the two triangles we just created have three pairs of congruent sides. Therefore, we know that they are congruent triangles according to the SSS Congruence Theorem. △ ABD ≅ △ CBD Let's identify all congruent corresponding parts.
Notice that ∠ ADB and ∠ CDB form a linear pair. According to the Linear Pair Postulate, these are supplementary angles. Since they are congruent, they must be right angles.
Conclusion
Now we see that the median is also an angle bisector, a perpendicular bisector, and the altitude of △ ABD all at once.
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Solution
Consider △ ABC.
With respect to the triangle, how many unique equations can you write using only the following numbers, symbols, and segments? Numbers:& $ 1/4 $ $ 1/3 $ $ 1/2 $ $ 2/3 $ [0.8em] Symbols:& $ - $ $ = $ [0.4em] Segments:& $ GE $ $ AE $ $ AG $ Do not count equations where terms are rearranged.
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According to the Centroid Theorem, the distance from the centroid of a triangle to a vertex is two thirds the distance from the vertex to the midpoint of the opposite side. Let's show this on our triangle.
Using this theorem, we can write our first two equations. 1. &$ AG $ $ = $ $ 2/3 $ $ AE $ [1em] 2. & $ GE $ $ = $ $ 1/3 $ $ AE $ We can also define AG and GE by subtracting AG from AE and GE from AE. That's another two equations. 3. &$ AG $ $ = $ $ AE $ $ - $ $ AG $ [1em] 4. &$ GE $ $ = $ $ AE $ $ - $ $ GE $ Also, notice that GE is two-thirds of AE, while AG is one-third of AE. Therefore, it must be that AF is half as long as GE. We can also show this algebraically.
Now we can write another equation. 5. &$ AG $ $ = $ $ 1/2 $ $ GE $ Finally, since AG is defined as 23 of AE and GE is 12 of AG, we can write a sixth equation. 6. & $ GE $ $ = $ $ 2/3 $ $ AE $ $ - $ $ 1/2 $ $ AG $
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More Theorems About Triangles
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