Euclid, a Greek mathematician, introduced the method of proving a geometrical result by using logical reasoning on previously proved and known results. He introduced basic concepts of Axioms, Theorems and Corollaries. He also introduced the five postulates about lines and points. He also introduced the definitions of points, lines, line segments, rays etc. In this section let us study about rays, line segments, parallel lines and perpendicular lines.

Line Segment: A line segment is a part of a line and it has two end points.
According to Euclid's Axiom, between two points there exists only one line segment.The following diagram shows the line segment drawn between two points.
Line Segment
In the above diagram, the line segments are $\overline{AB}$ and $\overline{PQ}$.
Also, $\overline{AB}$ = $\overline{BA}$
and $\overline{PQ}$ = $\overline{QP}$

Example: Name the five line segments in the following diagram.

Example of Line Segment
Ray is a part of a line, which has one end point. A ray can be extended indefinitely in only one direction. A ray has infinitely many points.
A ray is named by naming the initial point and naming a point in the direction of motion.

Ray
The above diagram shows the rays, $\overrightarrow{AB}$, $\overrightarrow{PQ}$, $\overrightarrow{XY}$

Example: Name the rays in the following diagrams.

Example of Ray
Parallel lines are those which do not meet any were in the plane when extended infinitely.

The distance between the parallel lines is always the same.
Real Life Examples:
The railway track, opposite edges of computer monitor, opposite edges of the rectangular table.

Parallel lines
In the above diagram, the line l is parallel to m, in symbol we denote $l\parallel m$.

Transversal: If a line intersects two or more parallel lines at distinct points, then it is called a transversal.

Parallel lines and angles

Transversal

In the above diagram, the transversal t, intersects the parallel lines at, P. Q and R.

Angles made by a transversal with the pair of parallel lines.

Pairs of Angles

In the above diagram, we can see that the transversal t cuts the parallel lines m and n.
The 8 angles formed are named as, $\angle 1$, $\angle 3$, $\angle 2$, $\angle 4$, $\angle 5$, $\angle 7$, $\angle 8$
We can classify these angles into the following pairs of angles with their respective properties.

1. Vertically opposite angles: Pair of vertically opposite angles are equal.
$\angle 1$ = $\angle 3$
$\angle 2$ = $\angle 4$
$\angle 5$ = $\angle 7$
$\angle 8$ = $\angle 8$

2. Linear Pair:
$\angle 1$ + $\angle 2$ = 180 o , $\angle 5$ + $\angle 6$ = 180 o
$\angle 2$ + $\angle 3$ = 180 o , $\angle 6$ + $\angle 7$ = 180 o
$\angle 3$ + $\angle 4$ = 180 o , $\angle 7$ + $\angle 8$ = 180 o
$\angle 4$ + $\angle 1$ = 180 o , $\angle 8$ + $\angle 5$ = 180 o

3. Corresponding Angles: Pair of corresponding angles are equal.
$\angle 1$ = $\angle 5$
$\angle 2$ = $\angle 6$
$\angle 3$ = $\angle 7$
$\angle 4$ = $\angle 8$

4. Alternate Interior angles: Pair of alternate interior angles are equal.
$\angle 3$ = $\angle 5$
$\angle 4$ = $\angle 6$

5. Alternate Exterior angles: Pair of alternate exterior angles are equal.
$\angle 2$ = $\angle 8$
$\angle 7$ = $\angle 7$

6. Co-interior angles: Pair of co-interior angles are supplementary. These angles are also called as consecutive interior angles.
$\angle 3$ + $\angle 6$ = 180 o
$\angle 4$ + $\angle 5$ = 180 o

Solved Example

Question: Find the measure of the angle x in the following diagram.
Example of Parallel Line

Solution:
 
In the above diagram, $GE\parallel AB$.
Therefore, $\angle GEA$ = $\angle EAB$ [ alternate interior angles ]
       =>   $\angle GEA$ = 40 o ----------------( 1 )
Moreover, $GE\parallel CD$.
Therefore, $\angle GEC$ = $\angle ECD$ [ alternate interior angles ]
      =>     $\angle GEC$ = 110 o -------------------( 2 )
         But $\angle GEC$ = $\angle GEA$ + $\angle AEG$
                                    = 40 + x   --------------------( 3 )
Therefore             40 + x = 110 [ equating ( 2 ) and ( 3 ) ]
      =>                        x = 110 - 40
                                     = 70 o
 

The pair of lines which intersect at 90 o is called perpendicular lines.The following diagram describe the pairs of perpendicular lines.
Perpendicular Lines
In symbols we can denote $l\perp m$ and $p\perp q$

Solved Examples

Question 1: In the following Rhombus, ABCD, find the value of x and y.
Rhombus Example

Solution:
 
We know that Rhombus is a parallelogram, where
1. Opposite sides are parallel.
2. Diagonals bisect the angles.
3. All sides measure same lengths
4. Diagonals bisect at right angles.

In the above Rhombus,
$\angle CAB$ = $\frac{1}{2}$ of $\angle DAB$ [ AC is the bisector of $\angle A$ ]

                     = $\frac{1}{2}$ of 80o

                     = 40 o
$\angle DCA$ = $\angle CAB$
                     = 40 o
Therefore, we have x = 40 o
Being a parallelogram, adjacent angles are supplementary
Therefore, $\angle DAB$ + $\angle ABC$ = 180 o
                  => 80          + $\angle ABC$ = 180 o
                  =>                  $\angle ABC$ = 180 o - 80 o
                                                             = 100 o
                                                          y = $\frac{1}{2}$ of $\angle ABC$

                                                             = $\frac{1}{2}$ of 100 o

                                                             = 50 o
Therefore, we have x = 40 o and y = 50 o
 

Question 2: Prove that the angular bisector of an equilateral triangle is perpendicular to the opposite base.
Example of Perpendicular Lines

Solution:
 
In $\Delta ABC$, AB = BC = CA. AD bisects $\angle A$
To prove : AD $\perp$ BC.
Proof:
                   
 Serial No.           Statement              
      Reason             
 1.  In $\Delta ABD$ and $\Delta ADC$,
 AB = AC
 $\Delta ABC$ is an equilateral triangle. 
 2.   $\angle BAD$ = $\angle CAD$  AD is the bisector of $\angle A$
 3.   AD = AD  Common side
 4.   $\Delta ABD$ $\cong$ $\Delta ADC$   SAS postulate
 5.   $\angle ADB$ = $\angle ADC$  congruent parts of congruent triangles
 6.   $\angle ADB$ + $\angle ADC$ = 180 o
 Linear Pair
 7.   $\angle ADC$ + $\angle ADC$ = 180 o   Substituting ( 5 ) in ( 6 )
 8.   2 $\angle ADC$ = 180 o   
 9.  $\angle ADC$ = $\frac{180}{2}$
                      = 90 o 
 dividing both sides by 2
 10.   Therefore, $\angle ADE$ = $\angle ADB$ = 90 o    from ( 5 )
 11.                     AD $\perp$ BC