Point of Concurrency
A carpenter designed a triangular table that had one leg. He used a special point of the table which was the center of gravity, due to which the table was balanced and stable.
Do you know what this special point is known as and how do you find it?
This special point is the point of concurrency of medians.
In this page, you will learn all about the point of concurrency.
This mini-lesson will also cover the point of concurrency of perpendicular bisectors, the point of concurrency of the angle bisectors of a triangle, and interesting practice questions.
Let’s begin!
Lesson Plan
What Is the Point of Concurrency?
The point of concurrency is a point where three or more lines or rays intersect with each other.
For example, referring to the image shown below, point A is the point of concurrency, and all the three rays l, m, n are concurrent rays.
Triangle Concurrency Points
Four different types of line segments can be drawn for a triangle.
Please refer to the following table for the above statement:
| Name of the line segment | Description | Example |
|---|---|---|
| Perpendicular Bisector | These are the perpendicular lines drawn to the sides of the triangle. |
|
| Angle Bisector | These lines bisect the angles of the triangle. |
|
| Median | These line segments connect any vertex of the triangle to the mid-point of the opposite side. |
|
| Altitude | These are the perpendicular lines drawn to the opposite side from the vertices of the triangle. |
|
As four different types of line segments can be drawn to a triangle, similarly we have four different points of concurrency in a triangle.
These concurrent points are referred to as different centers according to the lines meeting at that point.
The different points of concurrency in the triangle are:
- Circumcenter.
- Incenter.
- Centroid.
- Orthocenter.
1. Circumcenter
The circumcenter is the point of concurrency of the perpendicular bisectors of all the sides of a triangle.
For an obtuse-angled triangle, the circumcenter lies outside the triangle.
For a right-angled triangle, the circumcenter lies at the hypotenuse.
If we draw a circle taking a circumcenter as the center and touching the vertices of the triangle, we get a circle known as a circumcircle.
2. Incenter
The incenter is the point of concurrency of the angle bisectors of all the interior angles of the triangle.
In other words, the point where three angle bisectors of the angles of the triangle meet are known as the incenter.
The incenter always lies within the triangle.
The circle that is drawn taking the incenter as the center, is known as the incircle.
3. Centroid
The point where three medians of the triangle meet is known as the centroid.
In Physics, we use the term "center of mass" and it lies at the centroid of the triangle.
Centroid always lies within the triangle.
It always divides each median into segments in the ratio of 2:1.
4. Orthocenter
The point where three altitudes of the triangle meet is known as the orthocenter.
For an obtuse-angled triangle, the orthocenter lies outside the triangle.
Observe the different congruency points of a triangle with the following simulation:
- The circumcenter of an equilateral triangle divides the triangle into three equal parts if joined with each vertex.
- For an equilateral triangle, all the four points (circumcenter, incenter, orthocenter, and centroid) coincide.
- Any point on the perpendicular bisector of a line segment is equidistant from the two ends of the line segment.
Solved Examples
Let us see some solved examples to understand the concept better.
| Example 1 |
Ruth needs to identify the figure which accurately represents the formation of an orthocenter. Can you help her figure out this?
Solution
The point where the three altitudes of a triangle meet are known as the orthocenter.
Therefore, the orthocenter is a concurrent point of altitudes.
Hence,
| \(\therefore\) Figure C represents an orthocenter. |
| Example 2 |
Shemron has a cake that is shaped like an equilateral triangle of sides \(\sqrt3 \text { in}\) each. He wants to find out the radius of the circular base of the cylindrical box which will contain this cake.
Solution
Since it is an equilateral triangle, \( \text {AD}\) (perpendicular bisector) will go through the circumcenter \(\text O \).
The circumcenter will divide the equilateral triangle into three equal triangles if joined with the vertices.
So,
\[\begin{align*} \text {area} \triangle AOC &= \text {area} \triangle AOB = \text {area} \triangle BOC \end{align*}\]
Therefore,
\[\begin{align*} \text {area of } \triangle {ABC} &= 3 \times \text {area of } \triangle BOC \end{align*} \]
Using the formula for the area of an equilateral triangle \[\begin{align*} &= \dfrac{\sqrt3}{4} \times a^2 \hspace{3cm} ...1 \end{align*} \]
Also, area of triangle \[\begin{align*} &= \dfrac{1}{2} \times \text { base } \times \text { height } \hspace{1cm} ...2 \end{align*} \]
By applying equation 1 and 2 for \(\triangle \text{BOC}\) we get,
\[\begin{align*} {\dfrac{\sqrt3}{4}} \times a^2 &= 3\times \dfrac{1}{2} \times a\times OD\\OD &= \dfrac{1}{2{\sqrt3}} \times a \hspace{2cm} ...3\end{align*}\]
Now, by applying equation 1 and 2 for \(\triangle \text{ABC}\) we get,
\( \text{Area of the } \triangle \text{ ABC} \) \[= \dfrac{1}{2} \times \text { base } \times \text { height } = \dfrac{\sqrt3}{4} \times a^2 ...4 \]
Using equation 3 and 4, we get
\[\begin{align*}\dfrac {1}2\times a\times (R+OD) &= \dfrac {\sqrt 3}4\times a^2 \\\dfrac12 a\times \left( R+\dfrac a{2\sqrt3}\right) &= \dfrac{\sqrt3}4\times a^2\\R &= \dfrac a{\sqrt3} \end{align*}\]
substituting-
\[ \begin{align*}a & = \sqrt3 \end{align*}\]
|
\(\therefore\) \(\text {R} = 1 \text{ in}\) |
| Example 3 |
A teacher drew 3 medians of a triangle and asked his students to name the concurrent point of these three lines. Can you name it?
Solution
The point where three medians of the triangle meet are known as the centroid.
The concurrent point drawn by the teacher is-
| \(\therefore\) Centroid |
| Example 4 |
For an equilateral \(\triangle \text{ABC}\), if P is the orthocenter, find the value of \( \angle BAP\).
Solution
For an equilateral triangle, all the four points (circumcenter, incenter, orthocenter, and centroid) coincide.
Therefore, point P is also an incenter of this triangle.
Since this is an equilateral triangle in which all the angles are equal, the value of \( \angle BAC = 60^\circ\)
Hence, line AP is an angle bisector of the \(\angle BAC\). \[ \implies \angle BAP = \dfrac {\angle BAC}{2} = 30^\circ\]
| \(\therefore\) \( \angle BAP = 30^\circ \) |
- The centroid of a triangle cuts each median into two segments. The shorter segment is ___________ the length of the entire segment.
Interactive Questions
Here are a few activities for you to practice. Select/Type your answer and click the 'Check Answer' button to see the result.
Let's Summarize
We hope you enjoyed learning about the point of concurrency with the simulations and interactive questions. Now, you will be able to easily solve problems on point of concurrency of perpendicular bisectors, the point of concurrency of the angle bisectors of a triangle, and the point of concurrency of the perpendicular bisectors of a triangle.
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Frequently Asked Questions (FAQs)
1. What are the four common points of concurrency?
The four common points of concurrency are centroid, orthocenter, circumcenter, and incenter.
2. What point of concurrency in a triangle is always located inside the triangle?
The centroid and incenter of a triangle always lie inside a triangle.