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 of 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=32 AE,BG=32 BD,andCG=32 CF $
Consider a triangle with vertices $A,$ $B,$ and $C,$ and 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. 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=ECBC=BE+EC ⇒BC=2EC $ Therefore, the scale factor of the similar triangles is $21 .$ This also means that $DC=2QC.$ By the Segment Addition Postulate $DC=DQ+QC.$ 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. $ADDQ =DCDQ =21 $ Corresponding parts of similar triangles are proportional. Since $△AGD$ and $△AEQ$ are similar, the ratio of $DQ$ to $AD$ is equal to the ratio of $GE$ to $AG.$ $AGGE =21 ⇔GE=21 AG $ This information can be used to express $AG$ in terms of $AE.$$GE=21 AG$
$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 $AE.$ Therefore, $G$ and $K$ are the same point which means that the medians are concurrent, or meet at one point.
Before it was shown that $AG=32 AE.$ By using similar arguments it can be also shown that $BG=32 BD$ and $CG=32 CF.$
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?
$32 ⋅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.
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.
The circumcenter of a triangle is equidistant from each of the triangle's vertices.
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 $C.$
By the Transitive Property of Equality, $AP$ is equal to $CP.$ ${AS=BSBS=CS ⇒AS=CS $ This proves that $AS,$ $BS,$ and $CS$ are all equal.
The proof can be summarized in the following two-column table.
Statements | Reasons |
$ ABCis a triangleDSis a perpendicular bisector ofABESis a perpendicular bisector ofBCFSis a perpendicular bisector ofAC $ | Given |
$ AS=BSBS=CS $ | Perpendicular Bisector Theorem |
$AS=CS$ | Transitive Property of Equality |
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.
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=EIandEI=FI⇕DI=EI=FI $ This means that $I$ is equidistant from each of the triangle's sides.
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 passing through the vertices $A,B,$ and $C,$ and parallel to the opposite sides $BC,AC,$ and $AB,$ respectively.
Notice that $ABCD$ and $AEBC$ are parallelograms.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.
Statements | Reasons |
$ABC$ is a triangle | Given |
$ DE∥BCEF∥ACDF∥AB $ | Given (Drawn) |
$ABCD,AEBC,$ and $ABFC$ are parallelograms | Definition of a parallelogram |
$ BC≅AD,BC≅EAAC≅BE,AC≅FBAB≅DC,AB≅CF $ | Parallelogram Opposite Sides Theorem |
$ BC=AD,BC=EAAC=BE,AC=FBAB=CF,AB=DC $ | Definition of congruent segments |
$ AD=EABE=FBCF=DC $ | Transitive Property of Equality |
$ Ais the midpoint ofDEBis the midpoint ofEFCis the midpoint ofDF $ | Definition of a midpoint |
$AK,$ $BL, $ and $CM$ are the altitudes of $△ABC$ | Given |
$ AK⊥BCBL⊥ACCM⊥AB $ | Definition of an altitude |
$ AK⊥DEBL⊥EFCM⊥DF $ | Perpendicular Transversal Theorem |
$ AKis a perpendicular bisector ofDEBLis a perpendicular bisector ofEFCMis a perpendicular bisector ofDF $ | Definition of a perpendicular bisector |
$AK,$ $BL,$ and $CM$ are concurrent | Definition of circumcenter |
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?