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Proof Geometry 6-4: Perpendiculars.

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Presentation on theme: "Proof Geometry 6-4: Perpendiculars."— Presentation transcript:

1 Proof Geometry 6-4: Perpendiculars

2 The Angle Construction Postulate
H 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


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