9-1 and 9-2 Parallel lines Proof Geometry

Three ways in which two lines may be situated in space They may fail to intersect and fail to be coplanar: L1 and L3 Called skew They may intersect in a point: L1 and L2 In which case they are co-planar (Given two intersecting lines, there is exactly one plane containing both.) They may be coplanar without intersecting each other: L2 and L3 Called parallel

Definitions: Skew and Parallel Two lines are skew if the do not lie in the same plane Two lines are parallel if: They are coplanar They do not intersect We write On sketch

Theorem with words Exactly One … Two parallel lines lie in exactly one plane

Exactly one means: Existence: Uniqueness: We have to show that there exists at least one plane through the point. It could be more than one. All we know for sure at this point is that one exists. Uniqueness: We then have to show that there is at most one plane that exists. Uniqueness proofs are usually by contradiction. If we prove both existence and uniqueness we have proven that there is exactly one. (AKA one and only one)

What you need to do Understand and write a proof by contradiction Understand what uniqueness and existence mean (you do not need to be able to write these proofs). Understand math symbols used in logical reasoning. Understand and write a proof using an auxiliary set.

Existence The existence of the plane comes from the definition of parallel lines. Parallel lines are co-planar

Uniqueness Prove(by contradiction): There is one plane containing the parallel lines L1 and L2 Suppose: There are two planes E and E’ that contain L1 and L2(supposition) Then: If P is a point on L2 then plane E and E’ contain both the line L2 and the point P. (conclusion resulting from supposition) But: Earlier we had a Theorem saying that given a line and a point not on the line, there is exactly one plane containing both (the CONTRADICTION) So: There is only one plane that contains parallel lines L1 and L2

Lines perpendicular to same line In a plane, if two lines are both perpendicular to the same line, then they are parallel. then

Existence of Parallels Let L be a line and P be a point not on L. Then there is at least one line through P, parallel to L. then Existence can be shown by constructing a line perpendicular to L and then another perpendicular to this constructed line.

Parallel Postulate Parallel line unique? Cannot be proven (only assumed) Led to Euclid’s Fifth Postulate AKA Parallel Postulate: That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles. Video

Definition: Transversal A transversal of two coplanar lines is a line which intersects them in two different points. Transversal Not Transversal

Definition: Alternate Interior Angles Given two line L1 and L2, cut by transversal T at points P and Q. Let A be a point of L1 and let B be a point of L2, such that A and B lie on opposite sides of T. Then APQ and PQB are alternate interior angles. In both figures 1 and 2 are alternate interior angles

The Alternate Interior Parallel Theorem Given two lines cut by a transversal. If a pair of alternate interior angles are congruent, then the lines are parallel. then

Other Angles formed when two lines cut by transversal Corresponding Angles: Two angles that occupy corresponding positions. Same Side Interior Angles: Two angles that lie between parallel lines on the same sides of the transversal. 1 2 1 2 3 3 4 4 5 6 5 6 7 8 7 8

The Corresponding Angles Parallel Theorem Given two lines cut by a transversal. If a pair of corresponding angles are congruent, then the lines are parallel. If  2   6 or  1   5 or  3   7 or  4   8, then L1||L2 1 2 L1 3 4 5 6 L2 7 8

The Same Side Interior Angles Parallel Theorem Given two lines cut by a transversal. If a pair of interior angles on the same side of the transversal are supplementary angles, then the lines are parallel. If m3 +m5 = 180 or m4 +m6 = 180 then L1 || L2 1 2 L1 3 4 5 6 L2 7 8

Example Given: KMJ with KJ = MJ, GJ = HJ, and HGJ HMK Prove: GJ = HJ Given HGJ  JHG Isosc Triangle Thm HGJ  HMK Given JHG  HMK Transitive 2 and 3 Corresponding Angle Parallel Theorem

Homework pg. 266-268: # 2, 6, 10, 13 pg. 271-272: #1, 3, 5, 6 - 8