Concurrency
Now that we have learned about several types of segments, we need to talk about how they relate in a single triangle.
Three or more lines are concurrent if they come together and meet at a single point. The point at which three or more lines meet is called the point of concurrency. Take a look at the diagram below.
Line \(a\) and line \(b\) intersect at point \(E\). Lines \(a\), \( c\), and \(d\) intersect at point \(F\). Therefore point \(F\) is a point of concurrency.
Now that we have learned about several types of segments, we need to talk about how they relate in a single triangle.
Three or more lines are concurrent if they come together and meet at a single point. The point at which three or more lines meet is called the point of concurrency. Take a look at the diagram below.
Line \(a\) and line \(b\) intersect at point \(E\). Lines \(a\), \( c\), and \(d\) intersect at point \(F\). Therefore point \(F\) is a point of concurrency.
Let’s refer back to each of the previous targets to see how it connects with this target!
In Target A, we discussed midsegments. A single triangle has three midsegments, pictured to the right: Since all three midsegments don’t meet at one point (aka, they are not concurrent at one point), the three midsegments of a triangle do not have a point of concurrency. |
Circumcenter
In Target B, we discussed several segments. Let’s start with perpendicular bisectors. In a single triangle, each side has its own perpendicular bisector. As you take a look below, make sure to drag around the vertices to change the type of triangle.
In Target B, we discussed several segments. Let’s start with perpendicular bisectors. In a single triangle, each side has its own perpendicular bisector. As you take a look below, make sure to drag around the vertices to change the type of triangle.
Notice that the perpendicular bisectors of a triangle always meet at one point. The point of concurrency for the perpendicular bisectors of a triangle is called the circumcenter. The circumcenter is the equidistant to each vertex, and that is why we can draw a circle using the circumcenter as the center of the circle and the vertices of the triangle as points on the circle. Regardless of the type of triangle with which we start, this holds true. We can say that the circle is circumscribed about the triangle.
Incenter
Next, we will talk about angle bisectors. In a single triangle, each angle has its own angle bisector. As you take a look below, make sure to drag around the vertices to change the type of triangle.
Incenter
Next, we will talk about angle bisectors. In a single triangle, each angle has its own angle bisector. As you take a look below, make sure to drag around the vertices to change the type of triangle.
Notice that the angle bisectors of a triangle always meet at one point. The point of concurrency for the angle bisectors of a triangle is called the incenter. The incenter is the equidistant to the sides of the triangle (where it meets the sides at a right angle). For the incenter, we are able to draw a circle using the incenter as the center of the circle and the the sides of the triangle are tangent to the circle. We can say that the circle is inscribed in the triangle. Regardless of the type of triangle with which we start, this holds true. (See the summary table below)
Centroid
Now let’s talk about medians. In a single triangle, each side has its own median. As you take a look below, make sure to drag around the vertices to change the type of triangle.
Centroid
Now let’s talk about medians. In a single triangle, each side has its own median. As you take a look below, make sure to drag around the vertices to change the type of triangle.
Notice that the medians of a triangle always meet at one point. The point of concurrency for the medians of a triangle is called the centroid. Along a single median, the distance from the vertex to centroid is twice that of the distance from the centroid to the midpoint. Regardless of the type of triangle with which we start, this holds true. (See the summary box below)
Orthocenter
Last, let’s talk about the altitudes. In a single triangle, each side has its own altitude. Recall that an altitude must go from vertex to opposite side in a right angle. As you take a look below, make sure to drag around the vertices to change the type of triangle
Orthocenter
Last, let’s talk about the altitudes. In a single triangle, each side has its own altitude. Recall that an altitude must go from vertex to opposite side in a right angle. As you take a look below, make sure to drag around the vertices to change the type of triangle
Notice that the altitudes of a triangle always meet at one point. The point of concurrency for the altitudes of a triangle is called the orthocenter. (See the summary table below)
Segments, Lines or Rays |
Point of Concurrency (PoC) |
Theorem |
Acute Triangle Location of PoC |
Right Triangle Location of PoC |
Obtuse Triangle Location of PoC |
Perpendicular Bisectors |
Circumcenter |
Circumcenter is equidistant to the vertices of the triangle. |
Inside of the triangle |
On the triangle |
Outside of the triangle |
Angle Bisectors |
Incenter |
Incenter is equidistant to the sides of the triangle. |
Inside of the triangle |
Inside of the triangle |
Inside of the triangle |
Medians |
Centroid |
Distance from centroid to vertex is twice as long as the distance from centroid to midpoint on a single median |
Inside of the triangle |
Inside of the triangle |
Inside of the triangle |
Altitudes |
Orthocenter |
Inside of the triangle |
On the triangle |
Outside of the triangle |
Now that we know these theorems, let’s complete some examples that utilize them!
Example 1:
Given that \(D\) is the circumcenter of \(\triangle ABC\), \(AC = 30\) units, and \(DE = 8\) units, find \(DC\).
Example 1:
Given that \(D\) is the circumcenter of \(\triangle ABC\), \(AC = 30\) units, and \(DE = 8\) units, find \(DC\).
Solution:
Because \(D\) is the circumcenter, that means \(\overline{DE}\) , \(\overline{DG}\), and \(\overline{EF}\) are perpendicular bisectors. We also know that \(D\) is equidistant to the vertices of the triangle. Let’s mark the diagram to show what we know:
Because \(D\) is the circumcenter, that means \(\overline{DE}\) , \(\overline{DG}\), and \(\overline{EF}\) are perpendicular bisectors. We also know that \(D\) is equidistant to the vertices of the triangle. Let’s mark the diagram to show what we know:
Since \(\overline{DE}\) is a perpendicular bisector, that means that point \(E\) is a midpoint. Since \(AC = 30\) units, then \(AE = 15\) units and \(EC = 15\) units. We will choose to focus on \(\triangle AED\) since it is a right triangle where we know two side lengths:
So let’s use the Pythagorean Theorem:
\(\begin{align*}\\
(AE)^2 + (DE)^2 &= (AD)^2\\
(15)^2 + (8)^2 &= (AD)^2\\
255 + 64 &= (AD)^2\\
289 &= (AD)^2\\
\pm\sqrt{289} &= AD
\end{align*}\)
\(AD = 17\) units (length must be positive)
Recall that the circumcenter is equidistant to the vertices of the triangle. Since \(AD\), \(CD\) , and \(BD\) represent those distances to the vertices, they are all equal! Therefore, \(DC\) = 17 units.
Example 2:
Given point \(D\) is the incenter of \(\triangle ABC\), where \(DB = 10\) units, \(BE = 8\) units, and \(DE = 0.5x\). Find the value of \(x\).
\(\begin{align*}\\
(AE)^2 + (DE)^2 &= (AD)^2\\
(15)^2 + (8)^2 &= (AD)^2\\
255 + 64 &= (AD)^2\\
289 &= (AD)^2\\
\pm\sqrt{289} &= AD
\end{align*}\)
\(AD = 17\) units (length must be positive)
Recall that the circumcenter is equidistant to the vertices of the triangle. Since \(AD\), \(CD\) , and \(BD\) represent those distances to the vertices, they are all equal! Therefore, \(DC\) = 17 units.
Example 2:
Given point \(D\) is the incenter of \(\triangle ABC\), where \(DB = 10\) units, \(BE = 8\) units, and \(DE = 0.5x\). Find the value of \(x\).
Solution:
Because \(D\) is an incenter, let’s mark the diagram to show what we know:
Because \(D\) is an incenter, let’s mark the diagram to show what we know:
Let’s focus on \(\triangle DEB\), a right triangle where we know at least two side lengths. Let’s use the Pythagorean theorem to solve for \(x\):
\(\begin{align*}\\
(BE)^2 + (ED)^2 &= (DB)^2\\
(8)^2 + (0.5x)^2 &= (10)^2\\
64 + (0.5x)^2 &= 100\\
(0.5x)^2 &= 36\\
0.5x &= \pm\sqrt{36}
\end{align*}\)
This means that
\(0.5x = 6\) or \(0.5x = -6\)
\(x = 12\) or \(x = -12\) Divide both sides of the equations by \(0.5\)
Substituting both values in for \(x\): \(DE = 0.5x = 0.5(-12) = -6\) and \(DE = 0.5x = 0.5(12) = 6\). Only \(x = 12\) produces a positive length.
Example 3:
Use the coordinate plane and coordinate geometry to determine the coordinates of the centroid of \(\triangle ABC\) with \(A(-1, -3), B(0.4), C(7, 2)\).
Solution:
Let’s start by graphing \(\triangle ABC\):
\(\begin{align*}\\
(BE)^2 + (ED)^2 &= (DB)^2\\
(8)^2 + (0.5x)^2 &= (10)^2\\
64 + (0.5x)^2 &= 100\\
(0.5x)^2 &= 36\\
0.5x &= \pm\sqrt{36}
\end{align*}\)
This means that
\(0.5x = 6\) or \(0.5x = -6\)
\(x = 12\) or \(x = -12\) Divide both sides of the equations by \(0.5\)
Substituting both values in for \(x\): \(DE = 0.5x = 0.5(-12) = -6\) and \(DE = 0.5x = 0.5(12) = 6\). Only \(x = 12\) produces a positive length.
Example 3:
Use the coordinate plane and coordinate geometry to determine the coordinates of the centroid of \(\triangle ABC\) with \(A(-1, -3), B(0.4), C(7, 2)\).
Solution:
Let’s start by graphing \(\triangle ABC\):
In order to find the centroid of this triangle, we must use the midpoint formula to find the midpoints of each side:
Midpoint of \(\overline {AB}\): \(\left(\Large\frac{0 + -1}{2}, \frac{4 + -3}{2}\right) = \left(-\Large\frac{1}{2}, \frac{1}{2}\right)\)
Midpoint of \(\overline {AB}\): \(\left(\Large\frac{0 + 7}{2}, \frac{4 + 2}{2}\right) = \left(\Large\frac{7}{2}, \frac{6}{2}\right) = \left(\Large\frac{7}{2},\normalsize 3\right)\)
Midpoint of \(\overline {AB}\): \(\left(\Large\frac{-1 + 7}{2}, \frac{-3 + 2}{2}\right) = \left(\Large\frac{6}{2}, -\frac{1}{2}\right) = \left(3, -\Large\frac{1}{2}\right)\)
Let’s plot the midpoints now:
Midpoint of \(\overline {AB}\): \(\left(\Large\frac{0 + -1}{2}, \frac{4 + -3}{2}\right) = \left(-\Large\frac{1}{2}, \frac{1}{2}\right)\)
Midpoint of \(\overline {AB}\): \(\left(\Large\frac{0 + 7}{2}, \frac{4 + 2}{2}\right) = \left(\Large\frac{7}{2}, \frac{6}{2}\right) = \left(\Large\frac{7}{2},\normalsize 3\right)\)
Midpoint of \(\overline {AB}\): \(\left(\Large\frac{-1 + 7}{2}, \frac{-3 + 2}{2}\right) = \left(\Large\frac{6}{2}, -\frac{1}{2}\right) = \left(3, -\Large\frac{1}{2}\right)\)
Let’s plot the midpoints now:
To construct the centroid, we need to connect vertex to opposite midpoint and create all three medians in the triangle:
Since the centroid is the point of concurrency for the medians, our centroid happens to be where the three medians meet. The coordinates of the centroid is \((2, 1)\):
Quick Check
1) Given the diagram below, find the value of \(x\).
1) Given the diagram below, find the value of \(x\).
2) Given the diagram below and \(JA = x\) units, find the value of \(x\).
3) Use the coordinate plane and coordinate geometry to determine the coordinates of the circumcenter of with \(A(-1, 4)\), \(B(7, 0)\), and \(C(5, 4)\).
Quick Check Solutions
Quick Check Solutions