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.

You are watching: Point of concurrency of a triangle

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**

1. | What Is the Point of Concurrency? |

2. | Important Notes on the Point of Concurrency |

3. | Solved Examples on the Point of Concurrency |

4. | Challenging Questions on the Point of Concurrency |

5. | Interactive Questions on the Point of Concurrency |

## What Is the Point of Concurrency?

The point of concurrency is a**point **where three or more linesor raysintersect 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 atriangle.

Please refer to the following table for the above statement:

Name of the line segmentDescriptionExamplePerpendicular 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 theperpendicular 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 thecenter 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 theangle bisectors of all the interior anglesof thetriangle.

In other words, the point where three angle bisectorsof the angles of the triangle meet areknown 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 mediansof the triangle meet isknown as the centroid.

In Physics, we use the term"center of mass" and itlies 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 altitudesof the triangle meet isknown 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:

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 hasa cake that is shaped like an equilateral triangle of sides \(\sqrt3 \text { in}\) each. He wants to find out the radiusofthe circular base of the cylindricalbox which will contain this cake.

**Solution**

Since it isan 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 mediansof the triangle meet areknown 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.

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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 \( \angle BAC = 60^\circ\)