By Alfred S. Posamentier

*Advanced Euclidean Geometry* provides a radical evaluation of the necessities of high college geometry after which expands these techniques to complicated Euclidean geometry, to provide lecturers extra self belief in guiding pupil explorations and questions.

The textual content includes 1000s of illustrations created within the Geometer's Sketchpad Dynamic Geometry® software program. it really is packaged with a CD-ROM containing over a hundred interactive sketches utilizing Sketchpad™ (assumes that the consumer has entry to the program).

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**Example text**

A p p lication 1 Prove that the medians of a triangle are concurrent. • In AABQ ALy BMy and CN are medians (see Figure 2-7). Therefore AN = NB, BL = LCy and CM = MA. Multiplying these equalities gives us: (ANKBDiCM) = (NB){LC)(MA) or ^ ‘^ ~ = 1 NB LC MA INTERACTIVE 2-7 Drag vertices A, B, and Cto change the shape of the triangle and see that the medians always meet at one point. Thus by Ceva’s theorem, AL, BMy and CN are concurrent. < Again, it would be advisable to compare the conventional proof (that pre sented in the context of elementary geometry) for the concurrency of the alti tudes of a triangle to the following proof, which uses Cevas theorem.

We committed this mathemati cal sin when we divided by zero in the form of [{BC){AM) — (AB)(MN)], which was a consequence of the triangles proved to be similar earlier. 20 ADVANCED EUCLIDEAN GEOMETRY COMMON NOMENCLATURE Figure 1-31 illustrates some of the details we will consider in this book. We list them systematically now, with the general understanding that we may use a sym bol ambiguously when we can simplify our work without confusion. Thus we may use b to represent either a side of a triangle, its name, or its measure, as the context should make clear.

Lay off EB on BC equal in length to BD^ Connect point E to point F, the mid point of ADy and extend to meet AB at point G. Draw GD. Construct perpendicular bisectors of GD and GE. Because GD and ^ GE are not parallel, the per pendicular bisectors must meet at point K. Connect point К with points G, Dy £, and B. Because GK = KD and GK = KE {г point on the perpendicular bisector of a line segment is equidistant from the ends of a line seg ment), KD = KE. We con structed DB = EB. Therefore AKBD = AKBE (SSS) and mAKBD = mAKBE.