Perpendicular Bisector

6 Key excerpts on "Perpendicular Bisector"

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  eBook - ePub

  Introduction to Non-Euclidean Geometry

  • Harold E. Wolfe(Author)
  • 2013(Publication Date)
  • Dover Publications

    (Publisher)

  In Hyperbolic Geometry, as is the case in Euclidean, the Perpendicular Bisectors of the sides of a triangle are concurrent and so also are the bisectors of the angles, the altitudes and the medians. Here, however, the lines must at times be regarded as intersecting in ideal or ultra-ideal points. The proofs for these concurrence theorems are, on the whole, not so. easily obtained as in Euclidean Geometry. Some of the difficulty encountered is rather fundamental in character. We shall treat here only one of these theorems, one which we shall utilize very soon.

  Theorem. The Perpendicular Bisectors of the sides of a triangle are concurrent.

  There are three cases to be considered.

  CASE I. If the Perpendicular Bisectors of two of the sides of a a triangle intersect in an ordinary point, then it can be proved by congruence theorems, just as in Euclidean Geometry, that the Perpendicular Bisector of the third side passes through this point.

  CASE II. If the Perpendicular Bisectors of two of the sides of a triangle are non-intersecting, then they will have a common perpendicular and intersect in an ultra-ideal point. We shall prove that the Perpendicular Bisector of the third side is perpendicular to this common perpendicular also, that is, that it passes through the same ultra-ideal point.

    Figure 43

  Let ABC (Fig. 43 ) be the triangle, with A′, B′, C′ the midpoints of the sides opposite A, B, C, respectively. Let the perpendiculars to sides AB and BC at C′, and A′ be non-intersecting. They then have a common perpendicular MN. We are to prove that the perpendicular to AC at B′ is also perpendicular to MN.

  Construct AH, BJ, CK and B′L perpendicular to MN. Then, if lines AN, BN, BM and CM are drawn, it is easy to prove that AH and CK are each equal to BJ and hence equal to one another. Thus AHKC is a Saccheri Quadrilateral and the line B′L through B′, the midpoint of the summit, perpendicular to the base, is perpendicular to the summit also. This proves that the Perpendicular Bisectors of the three sides of the triangle have in this case an ultra-ideal point in common.

  CASE

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  Norm Derivatives And Characterizations Of Inner Product Spaces

  • Claudi Alsina, Justyna Sikorska, M Santos Tomas(Authors)
  • 2009(Publication Date)
  • World Scientific

    (Publisher)

  Chapter 4 Perpendicular Bisectors in Normed Spaces The center of the circumscribed circumference of a triangle is the intersec-tion point of the three Perpendicular Bisectors of its three sides. This is a classical situation in an i.p.s. What happens in a general normed linear space where various notions of orthogonality lead to alternative expressions for Perpendicular Bisectors? We explore such questions in this chapter. 4.1 Definitions and basic properties In a real i.p.s. ( X, angbracketleft· , ·angbracketright ) with dimension greater than or equal to 2, given two linearly independent vectors x, y ∈ X , we define the Perpendicular Bisector of the linear segment [ x, y ] = { αx + (1 − α ) y : 0 ≤ α ≤ 1 } as the set braceleftbigg x + y 2 + λu : λ ∈ R bracerightbigg , where u negationslash = 0 is a vector in the subspace generated by x and y (denoted by span( x, y )) and orthogonal to x − y . If u := αx + βy , then by the requirement angbracketleft x − y, u angbracketright = 0, we have α ( bardbl x bardbl 2 − angbracketleft y, x angbracketright ) = β ( bardbl y bardbl 2 − angbracketleft x, y angbracketright ) , and we may take u = ( bardbl y bardbl 2 − angbracketleft x, y angbracketright ) x + ( bardbl x bardbl 2 − angbracketleft y, x angbracketright ) y. 103 104 Norm Derivatives and Characterizations of Inner Product Spaces x y u x + y 2 + λu Figure 4.1.1 In a real normed linear space ( X, bardbl·bardbl ), we can replace the inner product angbracketleft x, y angbracketright by its generalizations ρ ′ ± ( x, y ). Namely, we consider the vectors u ± ( x, y ) = ( bardbl y bardbl 2 − ρ ′ ± ( x, y )) x + ( bardbl x bardbl 2 − ρ ′ ± ( y, x )) y (4.1.1) for all x, y in X , and define M ± ( x, y ) := braceleftbigg x + y 2 + λu ± ( x, y ) : λ ∈ R bracerightbigg . (4.1.2) It follows from the above that if x and y are linearly dependent, e.g. y = λx for some λ ∈ R , then M ± ( x, λx ) = x + λx 2 .

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  Elementary Geometry for College Students

  • Daniel C. Alexander, Geralyn M. Koeberlein, , , Daniel C. Alexander, Geralyn M. Koeberlein(Authors)
  • 2014(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  1.6.4 The Perpendicular Bisector of a line segment is unique. 1.7.1 If two lines meet to form a right angle, then these lines are perpendicular. 1.7.2 If two angles are complementary to the same angle (or to congruent angles), then these angles are congruent. 1.7.3 If two angles are supplementary to the same angle (or to congruent angles), then these angles are congruent. 1.7.4 Any two right angles are congruent. 1.7.5 If the exterior sides of two adjacent acute angles form perpendicular rays, then these angles are complementary. 1.7.6 If the exterior sides of two adjacent angles form a straight line, then these angles are supplementary. 1.7.7 If two line segments are congruent, then their midpoints separate these segments into four congruent segments. 1.7.8 If two angles are congruent, then their bisectors separate these angles into four congruent angles. 2.1.1 From a point not on a given line, there is exactly one line perpendicular to the given line. 2.1.2 If two parallel lines are cut by a transversal, then the pairs of alternate interior angles are congruent. 2.1.3 If two parallel lines are cut by a transversal, then the pairs of alternate exterior angles are congruent. 2.1.4 If two parallel lines are cut by a transversal, then the pairs of interior angles on the same side of the transversal are supplementary. 2.1.5 If two parallel lines are cut by a transversal, then the pairs of exterior angles on the same side of the transversal are supplementary. 2.3.1 If two lines are cut by a transversal so that two corresponding angles are congruent, then these lines are parallel. 2.3.2 If two lines are cut by a transversal so that two alternate interior angles are congruent, then these lines are parallel. 2.3.3 If two lines are cut by a transversal so that two alternate exterior angles are congruent, then these lines are parallel.

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  Introduction to Non-Euclidean Geometry

  • EISENREICH(Author)
  • 2014(Publication Date)
  • Academic Press

    (Publisher)

  8. Perpendicular BisectorS OF THE SIDES OF A TRIANGLE We have already noted the following two facts concerning the perpendic-ular bisectors of the sides of a triangle : 1. If two Perpendicular Bisectors of the sides of a triangle meet, then the third goes through their point of intersection and the three bisectors are therefore concurrent (III, §14, Ex. 6). 2. If two Perpendicular Bisectors of the sides of a triangle are parallel, with a common perpendicular, then all three bisectors are parallel, with this same common perpendicular (III, §14, Ex. 7). Also, we have noted that each of these cases actually occurs (III, §14, Ex. 8 and proof of Theo. 56). The remaining possibility is that some of the bisectors might be boundary parallels. This case occurs, too (§6, Ex. 4), and the state-ment for it is as follows : 3. If two Perpendicular Bisectors of the sides of a triangle are boundary parallels, then the third bisector is a boundary parallel to each of them. This statement is an immediate consequence of statements 1 and 2 above. We leave the proof as Exercise 1. It will be noted that statement 3 says nothing about directions of paral-lelism. If, for example, the two given bisectors g, h are parallel in the direction a, statement 3 does not say whether the third bisector i is parallel to g and h in this same direction (Fig. IV. 32) or whether it is parallel to them in dif-ferent directions ß and y (Fig. IV, 33). There are no other possibilities. We Fig. IV, 32 114 IV. PARALLELS WITHOUT A COMMON PERPENDICULAR Fig. IV, 33 shall now prove that the situation in Fig. IV, 33 cannot occur, and hence that 4. If the three Perpendicular Bisectors of the sides of a triangle are boundary parallels to each other, they are all parallel in the same direction. We start the proof by establishing a simple fact about the Perpendicular Bisectors for any triangle: There is always a line which intersects all three bisectors.

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  Topics in Geometry

  • Robert Bix(Author)
  • 2014(Publication Date)
  • Academic Press

    (Publisher)

  Proof: Let / and m be the two given lines, and let p and q be the angle bisectors. One pair of vertical angles is divided by p into angles of equal measure 0, and the other pair of vertical angles is divided by q into angles of equal measure y/. The adjacent angles formed by / and m have measures 20 and 2y/, and so we have 20 + 2y/ = 180° (by Theorem 0.6). Dividing by 2 gives 0 + if/ = 90°. Since p and q form the angle 0 + y/, p and q are perpendicular. • We want to prove that the angle bisectors at the vertices of a triangle lie by threes on common points. The proof is analogous to the proof in Section 4 that the Perpendicular Bisectors of the three sides of a triangle meet at a point. By Theorem 4.2, the Perpendicular Bisector of two points A and B consists of exactly those points equidistant from A and B. Similarly, we show that the angle bisectors of two lines / and m consist of exactly those points equidistant from / and m. We use the following definition. DEFINITION 5.4. In the Euclidean plane, let P be a point and let / be a line. The foot of the perpendicular from P to / is the point F where / intersects the line on P perpendicular to /. The distance from P to / is the distance between the points P and F (Figure 5.8). • The next result is analogous to Theorem 4.2. THEOREM 5.5. In the Euclidean plane, let / and m be two lines on a point O. Then the angle bisectors consist of exactly those points equidistant from / and m (Figure 5.8). Proof: Let P be a point, and let F and G be the feet of the perpendiculars from P to I and m. First, assume that P doesn't lie on / or m and that neither F nor G equals O (Figure 5.8). P is equidistant from / and m if and only if it is equidistant from F and G (by Definition 5.4). FOP and GOP are right triangles that have a common side OP. Thus, P is equidistant from F and G if and only if triangles FOP and GOP are congruent (by the Pythagorean Theorem 0.9 and the SSS Property 0.1).

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  Geometry: 1001 Practice Problems For Dummies (+ Free Online Practice)

  • Allen Ma, Amber Kuang(Authors)
  • 2022(Publication Date)
  • For Dummies

    (Publisher)

  10. AF CF A median of a triangle is a line segment that connects the vertex of a triangle to the midpoint of the opposite side. 11. ∠ ADB and ∠ CDB are right angles. The altitude of a triangle is a segment that connects the vertex of a triangle perpendicular to the opposite side. 12. ∠ ∠ ABE CBE ≅ A bisector divides an angle into two congruent angles. 13. AF CF A midpoint divides a segment into two congruent segments. 14. Right A Perpendicular Bisector divides a segment into two congruent segments and also forms right angles at the point of intersection. 15. 12.5 A midpoint divides a segment into two congruent segments. Set the two segments equal and solve for x: CE BE x x x 2 25 50 2 2 5 12 5 . 16. 80° A bisector divides an angle into two congruent angles. Set the two angles equal and solve for x: m ADE m CDE x x x x x ∠ ∠ = -= + -= = = 3 5 25 2 5 25 2 30 15 CHAPTER 18 Answers and Explanations 159 ANSWERS 1–100 After finding the value of x, plug it in to find the value of the desired angle: m ADE m ADC m ADE ∠ ∠ ∠ = -= = = ( ) = 3 15 5 40 2 2 40 80 ( ) ( )    17. 22 Perpendicular lines form right angles. Set the angle equal to 90° and solve for x: m BAD x x x ∠ = -= = = 90 5 2 0 9 0 5 110 22  18. 21 A bisector divides a segment into two congruent segments. This means that BE CE . Find the sum of BE and CE, set it equal to BC, and solve for x: BE CE BC x x x x x x x x ( ) ( ) 12 12 5 3 2 24 5 3 24 3 3 27 3 9 Now plug in the value of x to find BE: BE CE x 12 9 12 21 19. Linear pair A linear pair is a pair of adjacent angles whose sum is a straight angle. 20. Perpendicular Perpendicular lines are two lines that intersect to form right angles. 21. Acute An acute angle is an angle measuring more than 0° but less than 90°. 22. Isosceles An isosceles triangle is a triangle that has two congruent angles with the sides opposite them congruent as well.

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