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Line L is parallel to line n.

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Q: What- line m is perpendicular to line L, line n is perpendicular to line m.?

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N=l-m

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Let be a set of lines in the plane. A line k is transversal of if # , and # for all . Let be transversal to m and n at points A and B, respectively. We say that each of the angles of intersection of and m and of and n has a transversal side in and a non-transversal side not contained in . Definition:An angle of intersection of m and k and one of n and k are alternate interior angles if their transversal sides are opposite directed and intersecting, and if their non-transversal sides lie on opposite sides of . Two of these angles are corresponding angles if their transversal sides have like directions and their non-transversal sides lie on the same side of . Definition: If k and are lines so that , we shall call these lines non-intersecting. We want to reserve the word parallel for later. Theorem 9.1:[Alternate Interior Angle Theorem] If two lines cut by a transversal have a pair of congruent alternate interior angles, then the two lines are non-intersecting.Figure 10.1: Alternate interior anglesProof: Let m and n be two lines cut by the transversal . Let the points of intersection be B and B', respectively. Choose a point A on m on one side of , and choose on the same side of as A. Likewise, choose on the opposite side of from A. Choose on the same side of as C. Hence, it is on the opposite side of from A', by the Plane Separation Axiom. We are given that . Assume that the lines m and n are not non-intersecting; i.e., they have a nonempty intersection. Let us denote this point of intersection by D. D is on one side of , so by changing the labeling, if necessary, we may assume that D lies on the same side of as C and C'. By Congruence Axiom 1 there is a unique point so that . Since, (by Axiom C-2), we may apply the SAS Axiom to prove thatFrom the definition of congruent triangles, it follows that . Now, the supplement of is congruent to the supplement of , by Proposition 8.5. The supplement of is and . Therefore, is congruent to the supplement of . Since the angles share a side, they are themselves supplementary. Thus, and we have shown that or that is more that one point, contradicting Proposition 6.1. Thus, mand n must be non-intersecting. Corollary 1: If m and n are distinct lines both perpendicular to the line , then m and n are non-intersecting. Proof: is the transversal to m and n. The alternate interior angles are right angles. By Proposition 8.14 all right angles are congruent, so the Alternate Interior Angle Theorem applies. m and n are non-intersecting. Corollary 2: If P is a point not on , then the perpendicular dropped from P to is unique. Proof: Assume that m is a perpendicular to through P, intersecting at Q. If n is another perpendicular to through P intersecting at R, then m and n are two distinct lines perpendicular to . By the above corollary, they are non-intersecting, but each contains P. Thus, the second line cannot be distinct, and the perpendicular is unique. The point at which this perpendicular intersects the line , is called the foot of the perpendicular

L-1 electron configuration

Where m and n are statements m n is called the _____ of m and n.

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