In geometry, initially we learn about the fundamental concepts of point, line, line segments, ray, angles, types of angles, plane figures etc. When we learn about lines, we also learn about the Euclidean axioms on lines, points and lines. One of the axiom is only one line can be drawn through two distinct points in a plane. also we are familiar with the property that two non-parallel lines will intersect at one and only one point. In this section let us learn about the concurrent lines and the problems based on it.

Three or more lines in a plane are said to be concurrent, if all of them intersect at the same point. In the following diagram, the lines, l, m and n all intersect at the point P. So they are concurrent.

Concurrent Lines

Solved Examples

Question 1: In the figure given below, the lines l, m and n intersect at O. Find the values of x, y and z, it is being given that x : y : z = 2 : 3 : 5.

Concurrent Line
Solution:
 
Let us assume the other three angles as, a, b and c
We know that, when two lines intersect, vertically opposite angles are equal.
Therefore, a = x, b = y and c = z
Since the sum of all the angles around a point is 360o ,
                    we have a + y +c +  x + b + z  = 360 o
Substituting , a = x, b = y and c = z, we have
                                 x + y + z + x + y + z = 360 o
                                 =>        2x + 2y + 2z = 360o
                                 =>       2 ( x + y + z ) = 360o

                                 =>              x + y + z = $\frac{360}{2}$
   
                                                                  = 1800  ------------------------ ( 1 )
          We are given that x : y : z = 2 : 3 : 5,
                  Let x = 2k, y = 3k and z = 5k
Substituting this in equation ( 1 ), we get
                                              2k + 3k + 5k = 180
                                =>                      10 k = 180

                                =>                          k = $\frac{180}{10}$
                
                                                                = 18
Substituting k = 18 in x, y and z, we get,
                                         x = 2k = 2 ( 18 ) = 36o
                                         y = 3 k = 3 ( 18 ) = 54o
                                         z = 5 k = 5 ( 18 ) = 90o
 

Question 2: In the following figure, find the measure of the angles, x , y and z.
Concurrent Example
Solution:
 
In the following figure AB and CD intersect at O.
Therefore,      $\angle COA$ + $\angle AOD$ = 180o , [ linear pair ]
                              3x - 20 +        x             = 180o
                      =>               4x - 20               = 180
                      =>                      4 x              = 180 + 20
                                                                  = 200

                      =>                         x             = $\frac{200}{4}$
                       
                                                                 = 50o
                                            $\angle AOC$ = 3x - 20
                                                                 = 3 ( 50 ) - 20
                                                                 = 150 - 20
                                                                 = 130o
                                            $\angle AOD$ = x = 50o
                                           $\angle BOC$ = y = x [ vertically opposite angles ]
                                                                = 50o
                                           $\angle BOD$ = z = 3x - 20  [ vertically opposite angles ]
                                                                = 130o
Therefore, x= 50o , y = 50o and z = 130o
 

What are Medians? Medians are the lines joining the vertex to the midpoint of the opposite sides.

Medians of any triangle are concurrent and the point of intersection is called the centroid.
The centroid of a triangle divides the median in the ratio 2 : 1.

The following diagram shows the point of intersection of the median.s of a triangle.

Triangle Median


The Medians of any triangle meet in the interior of the a triangle.
In the above triangle, AG : GD = 2 : 1. BG : GE = 2 : 1. CG : GF = 2 : 1

Altitudes:

Altitudes are the perpendiculars drawn from the vertex to the opposite sides.
The altitudes of any triangle are concurrent and the point of intersection of the altitudes is called an Orthocenter.

The following diagram shows the point of intersection of the altitudes of a triangle.

Triangle Altitude
1. The above triangle is an acute angled triangle and we see that the orthocenter lies inside the triangle.

2. For an obtuse angled triangle the orthocenter lies outside the triangle.

3. For a right triangle, the orthocenter is the vertex containing the right angle.

In any triangles, the altitudes, medians, and internal bisectors of the vertices or perpendicular bisectors of the sides are concurrent.

Let us see the special names for these points of concurrence.

Orthocenter: It is the point of concurrence of the altitudes of a triangle.

The following diagram shows the position of orthocenter for the respective triangle.

Orthocenter
1. For an acute angled triangle the orthocenter lies inside the triangle.
2. For an obtuse angled triangle the orthocenter lies outside the triangle
3. For a right angled triangle orthocenter lies at the hypotenuse of the triangle.

Centroid: It is the point of concurrence of the medians of a triangle.

Whether the triangle is acute, obtuse of right angled triangle, the centroid always lies inside the triangle.

Triangle Median

Incenter: It is the point of concurrence of the internal bisectors of the angles of a triangle.


Incenter
The circle with center as I and radius equal to the perpendicular distances of I from the sides, touches the sides of the triangle. Hence the point is called Incenter of the triangle.
For all types of triangles, acute, obtuse or right triangle, the Incenter always lies inside the circle.

Circumcenter: It is the point of concurrence of the perpendicular bisectors of the sides of a triangle.


Circumcenter
SA = SB = SC
We can see that the circle with center at S (circumcenter) is equidistant from the vertices and hence the circle passes through the vertices of the triangle. The circle so drawn is called as circumcircle.
1. For an acute angled triangle, the circumcenter lies in the interior of the triangle.
2. For a right angled triangle the circumcenter is the midpoint of the hypotenuse.
3. For an obtuse angled triangle, the circumcenter will lie outside the triangle.

Equilateral triangle: For an equilateral triangle, the above four points coincide. That is the points G, O, I and S are the same.


Equilateral Triangle Concurrency