|
Also see "Perpendicular Bisectors in a Triangle" at Segments in Triangles.
|
A perpendicular bisector of a given line segment is a line (or segment or ray) which is perpendicular to the given segment and intersects the given segment at its midpoint (thus "bisecting" the segment). |
|


|
The perpendicular bisector of a line segment is the set of all points that are equidistant from its endpoints. To be discussed further in the section on Constructions. |
|

Every point on the perpendicular bisector, , is the same distance from point A as it is from point B.
AC = CB
AF = FB
AG = GB
AE = EB
AH = HB
AJ = JB
AD = DB
You can think of the points on the perpendicular bisector as being the third vertices of a series of isosceles triangles with vertices A and B. |
|

Theorem Proof: (transformational method) |
|
|
1. A perpendicular bisector of a segment (by definition) is a line that is perpendicular to the segment and intersects the segment at its midpoint.
|
2.  because a midpoint of a segment divides the segment into two congruent segments.
|
3. AD = DB because congruent segments are segments of equal measure.
|
4. The line of reflection for a segment is the perpendicular bisector of the segment.
|
5. Under a reflection in  , A is mapped onto B, C is mapped onto C, and D is mapped onto D.
|
6. Under a reflection in  ,  is mapped onto  .
|
7. CA = CB since a reflection is a rigid transformation which preserves length. |

Theorem Proof: (two-column method) |
Statements |
Reasons |
1.  |
1. Given |
2.  |
2. Segment bisector forms 2 congruent segments. |
3. ∠ADC, ∠BDC are right angles |
3. Perpendiculars form right angles. |
4. ∠ADC ∠BDC |
4. All right angles are congruent. |
5.  |
5. Reflexive property |
6.  |
6. SAS: If 2 sides and the included angle of one triangle are congruent to the corresponding parts of another triangle, the triangles are congruent. |
7.  |
7. CPCTC: Corresponding parts of congruent triangles are congruent. |
8. CA = CB |
8. Congruent segments have equal measure. |

NOTE: The re-posting of materials (in part or whole) from this site to the Internet
is copyright violation
and is not considered "fair use" for educators. Please read the "Terms of Use". |
|
|