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| 12 Theory slides |
| 9 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
As it can be seen, medians of a triangle meet at a point. This point has a special name.
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 ba is constant for each median.
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.
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.
GE=21AG
a=22⋅a
b1⋅a=ba
Add fractions
LHS⋅2=RHS⋅2
LHS/3=RHS/3
ca⋅b=ca⋅b
Rearrange equation
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=32AE. By using similar arguments, it can be also shown that BG=32BD and CG=32CF.
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.
See solution.
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?
The centroid is 4 feet away from the vertex.
Therefore, if Zosia places the leg at this point, the table will be perfectly balanced.
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.
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.
Based on the characteristics of the diagram, the following relation holds true.
AS=BS=CS
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.
The proof can be summarized in the following two-column table.
0. Statements
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0. Reasons
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1. ABC is a triangleDS is a perpendicular bisector of ABES is a perpendicular bisector of BCFS is a perpendicular bisector of AC
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1. Given
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2. AS=BSBS=CS
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2. Perpendicular Bisector Theorem
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3. AS=CS
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3. Transitive Property of Equality
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Through exploration of the applet, it has been shown that the angle bisectors of a triangle intersect at one point.
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.
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
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.
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?
Note that the candle should be equidistant from each corner of the table.
Since the snacks should be equidistant from each side of the table, begin by recalling the Incenter Theorem.
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. |
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.
Circumcenter Theorem |
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.
As could be found in the previous exploration, when the altitudes of a triangle are drawn, they intersect at one point.
The orthocenter of a triangle is the point where a triangle's altitudes intersect. It is usually denoted by the letter H.
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.
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.
The proof can be summarized in the following two-column table.
0. Statements
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0. Reasons
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1. ABC is a triangle
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1. Given
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2. DE∥BCEF∥ACDF∥AB
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2. Given (Drawn)
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3. ABCD,AEBC, and ABFC are parallelograms
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3. Definition of a parallelogram
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4. BC≅AD,BC≅EAAC≅BE,AC≅FBAB≅DC,AB≅CF
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4. Parallelogram Opposite Sides Theorem
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5. BC=AD,BC=EAAC=BE,AC=FBAB=CF,AB=DC
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5. Definition of congruent segments
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6. AD=EABE=FBCF=DC
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6. Transitive Property of Equality
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7. A is the midpoint of DEB is the midpoint of EFC is the midpoint of DF
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7. Definition of a midpoint
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8. AK, BL, and CM are the altitudes of △ABC
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8. Given
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9. AK⊥BCBL⊥ACCM⊥AB
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9. Definition of an altitude
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10. AK⊥DEBL⊥EFCM⊥DF
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10. Perpendicular Transversal Theorem
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11. AK is a perpendicular bisector of DEBL is a perpendicular bisector of EFCM is a perpendicular bisector of DF
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11. Definition of a perpendicular bisector
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12. AK, BL, and CM are concurrent
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12. Definition of circumcenter
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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?
Point E is the centroid of △ABC.
Find the value of DE if BD is 12 centimeters.
Find the value of BE if BD is 21 centimeters.
Find the value of BD if ED is 7 centimeters.
Let's recall the Centroid Theorem. In a triangle, the centroid is always two thirds of the distance from a vertex to the triangle's midpoint on the opposite side. Consequently, this means DE must be one third of BD.
Since we know that BD equals 12 centimeters, we can calculate DE by using the equation in the diagram.
As in the previous exercise, we will use the Centroid Theorem to find BE. Let's first highlight the length of BE, which is two thirds of BD. For good measure, we will also include the length of ED.
Let's substitute the length of BD into the equation that describes BE and evaluate.
As in the previous exercises, we will again use the Centroid Theorem to determine the length of BD. Let's identify the length of the segments that make up BD.
Let's substitute the length of ED into the equation that describes it and evaluate for BD.
Let's draw the triangle with the given coordinates.
Since the triangle is obtuse, the orthocenter must necessarily fall outside of the triangle.
The orthocenter is the point of concurrency for a triangle's altitude. An altitude of a triangle is the perpendicular segment from a vertex to the side opposite the vertex. It can also fall on a line that coincides with the opposite side.
Let's draw the altitudes of our triangle. Notice that we only need to draw two of them to locate the orthocenter. Perhaps the easiest one to draw is the altitude from A since its opposite side, BC, is horizontal. This must mean that the altitude from A is a vertical line. For the purpose of finding the orthocenter, we will make it a ray.
Now we need to find one more altitude. Let's draw the altitude from B. This means AC becomes the triangle's base. Since the altitude should be perpendicular to the base, we must find a line through B that is perpendicular to AC. For this purpose, we will measure the vertical and horizontal distance between A and C.
The slope from A to C is - 43. Since the product of the slopes of perpendicular lines is -1, we can calculate the slope of the altitude drawn from B.
If we travel 4 units to the right and 3 units up from B, we get a second point that is located on the altitude. We can connect these two lines to draw our second altitude.
The orthocenter is the point where the two altitudes meet. For this triangle, it is located at (0,-5).
P is the incenter of △ABC.
Find PF if PD=6x−2 and PE=3x+10.
Find AP if PD=2x−2 and PE=12+x.
According to the Incenter Theorem, a triangle's incenter is equidistant from the triangle's three sides.
Therefore, we can set PD and PE equal to each other and solve for x. Then we use this information to calculate either PD or PE. In doing so, we will also have found the length of PF.
Now that we know x, we can calculate either PD or PE and thereby also find PF, as all three segments are congruent.
The length of PD is 22 centimeters, which means that this is also the length of PF.
We will divide the solution into two parts. First, we will find the length of PD. Then, we will use this information to find the length of AP.
As in the previous exercise, we can use the Incenter Theorem to state that PE, PD, and PF are congruent segments.
Since PD and PE are equal, we will set these expressions equal to each other and solve for x.
Now that we know x, we can calculate PD.
Notice that △ ABC is an equilateral triangle. Therefore, the angle bisectors divide the opposite sides into two congruent segments, which also makes P the triangle's centroid. According to the Centroid Theorem, the centroid is always two-thirds the distance from a vertex to the triangle's midpoint on the opposite side.
Let's rewrite the two equations showing AP and PD such that AD is isolated in both equations. AP=2/3AD ⇔ 3AP/2= AD [1em] PD=1/3AD ⇔ 3PD= AD Now we can combine these equations and solve for AP. 3AP/2= 3PD ⇔ AP=2PD Since we know the length of PD, we can find the length of AD.
The length of AP is 52 centimeters.
Let's begin by drawing the triangle in a coordinate plane using the given coordinates.
To find the centroid of a triangle, we first should determine the midpoint of each side.
We can find the midpoints by using the Midpoint Formula.
Side | Endpoints | M(x_1+x_2/2,y_1+y_2/2) | Midpoint |
---|---|---|---|
AB | ( -3,3), ( 9,1) | D(-3+ 9/2,3+ 1/2) | D(3,2) |
BC | ( 9,1), ( 3,9) | E(9+ 3/2,1+ 9/2) | E(6,5) |
AC | ( -3,3), ( 3,9) | F(-3+ 3/2,3+ 9/2) | F(0,6) |
Let's add these midpoints to our graph.
According to the Centroid Theorem, the centroid of a triangle is two-thirds of the distance from a vertex to the midpoint of the opposite side. Notice that D(3,2) has the same x-coordinate as C. Therefore, we can calculate the distance as the difference between their y-coordinates multiplied by 23.
This calculation tells us that the centroid is 143 units south of C along the median. That means it is 143 units below C. C(3,9) → (3,9- 14/3) → (3,13/3) The coordinates of the triangle's centroid are (3, 133).
Complete the statement with always, sometimes, or never.
The centroid is ____________ on the triangle. |
The orthocenter is ____________ outside the triangle. |
The centroid of a triangle is the point of concurrency of the triangle's medians, segments drawn between a vertex and the midpoint of the opposite side.
Since the midpoint of a triangle's side always falls between two vertices, the median can never fall on the triangle's perimeter. It does not matter if the triangle is acute, right, or obtuse. It can never happen!
The orthocenter of a triangle is the point where a triangle's altitudes intersect. An altitude of a triangle is a line segment drawn from a vertex and perpendicular to the opposite side.
Note that an altitude of a triangle is not always inside the triangle. Consider an acute triangle, a right triangle, and an obtuse triangle.
As it can be seen, the orthocenter can be inside, on, and outside the triangle. Therefore, the missing word is sometimes.