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