Lesson A perpendicular bisector of a segment

Source code of 'A perpendicular bisector of a segment'

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<H2>A perpendicular bisector of a segment</H2>

In this lesson you will learn about the properties of a perpendicular bisector of a line segment.
The lesson is grounded on the lesson <A HREF=http://www.algebra.com/algebra/homework/Triangles/Congruence-tests-for-triangles.lesson> Congruence_tests_for_triangles</A> of the current topic Triangles, the section Geometry in this site.

Theorem 1 (Perpendicular bisector/endpoints theorem)

If a point lies on the perpendicular drawn through the midpoint of a line segment, then the point is equidistant from the endpoints of the line segment.

<TABLE>
 <TR>
 <TD>
Proof

We are given the line segment AB (Figure 1), its midpoint D and the point C,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
which lies at the perpendicular CD passing through the point D.
We need to prove that the line segments AC and BC joining the point C with 
the segment AB endpoints are of equal length.

The proof is very simple. In the triangles ADC and BDC the sides AD and BD 
are congruent by the condition, because the point D is the midpoint of the 
segment AB. The side DC is the common side. The angles ADC and BDC 
are congruent as the right angles. Hence, the triangles ADC and BDC are 
congruent in accordance to the postulate 1 (SAS) of the lesson <A HREF=http://www.algebra.com/algebra/homework/Triangles/Congruence-tests-for-triangles.lesson> 
Congruence_tests_for_triangles</A> of the current topic Triangles, the section 
Geometry in this site.
 </TD>
 <TD>
{{{drawing( 200, 200, 0, 4, 0, 4, 
 line( 0.3, 0.5, 3.7, 0.5), 
 line( 0.3, 0.5, 2.0, 3.5),
 line( 2.0, 3.5, 3.7, 0.5),

locate(0.3, 0.5, A),
            locate(3.7, 0.5, B),
            locate(2.0, 3.9, C),

red(line( 2.0, 3.5, 2.0, 0.5)),
            locate(2.0, 0.5, D),

arc (2.0, 0.5, 0.8, 0.8, 180, 270),
            arc (2.0, 0.5, 1.0, 1.0, 180, 270),

arc (2.0, 0.5, 0.8, 0.8, 270, 360),
            arc (2.0, 0.5, 1.0, 1.0, 270, 360),

line( 1.1, 0.6, 1.1, 0.4),
 line( 2.9, 0.6, 2.9, 0.4)
)}}}
Figure 1. To the Theorem 1
 </TD>
 </TR>
</TABLE>

Therefore, the segments AC and BC are of equal length as the corresponding sides of the congruent triangles.
This is what to be demonstrated.

The opposite statement to the Theorem 1 is valid too.

Theorem 2 (Converse of perpendicular bisector/endpoints theorem)

If a point is equidistant from the endpoints of a line segment, then the point lies on the perpendicular drawn through the midpoint of the line segment.

<TABLE>
 <TR>
 <TD>
Proof

This time we are given the line segment AB (Figure 2) and the point C,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
equidistant from its endpoints A and B: AC = BC.
We need to prove that the point C lies on the perpendicular drawn 
through the midpoint of the line segment.

Let D be the midpoint of the line segment AB. Connect the points C and D 
by the straight line CD (Figure 2). Consider the triangles ADC and BDC. 
These triangles have the common side CD, the congruent sides AD and BD 
and the congruent sides AD and BD. Hence, the triangles are congruent 
in accordance to the postulate 3 (SSS) of the lesson <A HREF=http://www.algebra.com/algebra/homework/Triangles/Congruence-tests-for-triangles.lesson> 
Congruence_tests_for_triangles</A> of the current topic Triangles, the section 
Geometry in this site. 
 </TD>
 <TD>
{{{drawing( 200, 200, 0, 4, 0, 4, 
 line( 0.3, 0.5, 3.7, 0.5), 
 line( 0.3, 0.5, 2.0, 3.5),
 line( 2.0, 3.5, 3.7, 0.5),

locate(0.3, 0.5, A),
            locate(3.7, 0.5, B),
            locate(2.0, 3.9, C),

red(line( 2.0, 3.5, 2.0, 0.5)),
            locate(2.0, 0.5, D),

arc (2.0, 0.5, 0.8, 0.8, 180, 270),
            arc (2.0, 0.5, 1.0, 1.0, 180, 270),

arc (2.0, 0.5, 0.8, 0.8, 270, 360),
            arc (2.0, 0.5, 1.0, 1.0, 270, 360),

line( 1.1, 0.6, 1.1, 0.4),
            line( 2.9, 0.6, 2.9, 0.4),

line( 1.1, 2.0,  1.3, 2.0),
            line( 1.1, 2.08, 1.3, 2.08),

line( 2.9, 2.0, 2.7, 2.0),
 line( 2.9, 2.08, 2.7, 2.08)
)}}}
Figure 2. To the Theorem 2
 </TD>
 </TR>
</TABLE>

Therefore, the angles ADC and BDC are congruent as the corresponding angles of the congruent triangles. 
Since the sum of these angles ADC and BDC is the straight angle ADC of 180�, each of angles ADC and BDC is half of 180�, that is 90�.
This means that the segment CD is perpendicular to AB. 
Thus, the point C lies on the perpendicular CD drawn through the midpoint D of the line segment AB. This is exactly what to be demonstrated.

Summary

A point is equidistant from the endpoints of a line segment if and only if the point lies on the perpendicular drawn through the midpoint of the line segment.

We can present this result in the following terms:

&nbsp;&nbsp;&nbsp;&nbsp;- the set of points equidistant from two given points is the straight line perpendicular to the line segment connecting these points and erected at the

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;middle point of the segment,

or 
 &nbsp;&nbsp;&nbsp;&nbsp;- the straight line perpendicular to the line segment connecting two given points and erected at the middle point of the segment is the set of points

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;equidistant from the given points.

<BLOCKQUOTE>Definition

In Geometry a locus is the set of points satisfying some certain condition.</BLOCKQUOTE>

Based on the results obtained in this lesson, one can say that 
 &nbsp;&nbsp;&nbsp;&nbsp;- the perpendicular drawn through the midpoint of a line segment is the locus of points equidistant from the endpoints of the segment, &nbsp;&nbsp;or

&nbsp;&nbsp;&nbsp;&nbsp;- the locus of points equidistant from the endpoints of a line segment is the perpendicular drawn through the midpoint of the segment.

For navigation over the lessons on Properties of Triangles use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Triangles/Compendium-of-properties-of-triangles.lesson>Properties of Trianles</A>.

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Triangles/GEOMETRY-your-online-textbook.lesson>GEOMETRY - YOUR ONLINE TEXTBOOK</A>.