1. Proof Geometry
6-4: Perpendiculars

2. The Angle Construction PostulateH H
mPAB=r

3. Basic Perpendicular Line Theorem (point on given line)
In a given plane, through a given point on a line, there exists exactly one perpendicular line.

4. Basic Perpendicular Line Theorem (point on given line)
In a given plane, through a given point on a line, there exists exactly one perpendicular line. Existence: Through the Angle Construction Postulate exists such that where X is a point on line L. Let M= . Then

5. Basic Perpendicular Line Theorem (point on given line)
In a given plane, through a given point on a line, there exists exactly one perpendicular line. Uniqueness: Through the Angle Construction Postulate and exists so that and . We need to show that and are the same line.

6. Basic Perpendicular Line Theorem (point on given line)
In a given plane, through a given point on a line, there exists exactly one perpendicular line. We need to show that and are the same line.
Suppose: and are not the same line (supposition)
Then: because both are perpendicular to L
(conclusion resulting from supposition)
But: and intersect in P
(the CONTRADICTION)
So: and have to be the same line.

7. Perpendicular bisector Definition
In a plane, the perpendicular bisector of a segment is the line which is perpendicular to the segment at its midpoint.

8. Called a characterization theorem.
The Perpendicular Bisector Theorem: In a given plane, the perpendicular bisector of a segment is the set of all points that are equidistant from the endpoints of the segment.
Called a characterization theorem.
To characterize a set of points, we state a condition which:
1) Is satisfied by the points of a given set
2) Is not satisfied by any other points
In this case the set is the perpendicular bisector of and the condition is PA = PB

9. The Perpendicular Bisector Theorem: In a given plane, the perpendicular bisector of a segment is the set of all points that are equidistant from the endpoints of the segment. is the perpendicular bisector of .

10. The Perpendicular Bisector Theorem:
L is the perpendicular bisector of
A. Prove: If P is on L then PA = PB
case 1: P=C then PA = PB
because C is midpoint of
case 2: if P C but on L, then:
PC = PC Reflexive
2. CA = CB C is midpt.
3. PCA  PCB Both are right s
4. PAC  PBC SAS
5. PA = PB CPCTC

11. The Perpendicular Bisector Theorem:
L is the perpendicular bisector of
B. Prove: If PA = PB then P is on L.
case 1: If P is on , then P=C
Because has only one midpoint.
case 2: If P is not on , then:
PA = PB Given
CA = CB C is midpt.
PC =PC Reflexive
PAC  PBC SSS
PCA  PCB CPCTC
mPCA +mPCB=180 Supplement Postulate
mPCA = 90, mPCB=90 Cong. s in linear pair

12. Corollary 6-2. 1: Given a segment 𝐴𝐵 and a line L in a plane
Corollary 6-2.1: Given a segment 𝐴𝐵 and a line L in a plane. If two points of L are equidistant from A and B, then L is the perpendicular bisector of 𝐴𝐵 . Picture:
Each point that is equidistant from A and B are on L.
Since two points determine a line it means that L is the
perpendicular bisector.

13. Basic Perpendicular Line Theorem (point not on given line)
Given a line L and a point P not on the line, there exists exactly one line perpendicular to L that contains P.
We can show existence (at least one line perpendicular to L through point P) using congruent triangles.
We can show uniqueness (at most one line perpendicular to L through point P) using an indirect proof.
We will not study these proofs in detail.

14. Homework
pg : # 3-7, 11-13, 15, 16