Advanced Euclidean Geometry by Alfred S. Posamentier

By Alfred S. Posamentier

Advanced Euclidean Geometry provides an intensive overview of the necessities of high college geometry  after which expands these strategies to complicated Euclidean geometry, to provide academics extra self assurance in guiding scholar explorations and questions.

The textual content comprises hundreds of thousands of illustrations created within the Geometer's Sketchpad Dynamic Geometry® software program. it truly is packaged with a CD-ROM containing over a hundred interactive sketches utilizing Sketchpad™ (assumes that the person has entry to the program).

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Because these are the lines joining the corresponding vertices of AABC and ALMN, by Desargues s theorem the intersections of the corresponding sides are collinear; therefore points P, Q, and P are collinear. • Opplic^tion 7 Q roof In AABCy points F, Ey and D are the feet of the altitudes drawn from the vertices A, B, and C, respectively. The sides of p e d ^ triangle FEDy EFy DP, and DP intersect ^ ^ ^the sides of AABC, AB, AC, and BC, at points M, N, and L, respectively. Prove that points My Ny and L are collinear (see Figure 3-12).

Q roof In AABC, BM and CN are the interior angle bisectors, while AL bisects the exterior angle at point A (see Figure 3-5). Because the bisector of an angle (inte­ rior or exterior) of a triangle partitions the opposite side proportionally to the two remaining sides, we have: MC ~ BC N A ~ AC CL _ AC BL ~ AB Therefore, by multiplication: AM BN O L _ M ^ MC* NA^ B L ~ BC^ A c ' A B ~ However, because ^ = —— : BL LB AM BN CL _ MC* N A* LB ~ Thus, by Menelaus's theorem, points N, M, and L must be collinear.

You can follow this proof for both diagrams of Figure 2-8 because the same proof holds true for both an acute and an obtuse triangle. AANC ~ AAMB = ABLA ~ A B N C : AN AC M A ~ AB ^ N B ~ BC CM BC ACMB ~ ACLA = LC ~ AC (I) (II) (III) Chapter 2 33 CONCURRENCY of LINES in a TRIANGLE INTERACTIVE 2-8 Drag vertices A, B, and (7 to change the shape of the triangle and see that the altitudes always meet at one point. FIGURE 2-8 Multiplying (I), (II), and (III) gives us: m a ’n b ' lc ~ a b ' bc ‘AC~ ^ Therefore the altitudes are concurrent (by Cevas theorem).

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