By Alfred S. Posamentier

*Advanced Euclidean Geometry* provides an intensive evaluate of the necessities of high university geometry after which expands these suggestions to complex Euclidean geometry, to offer academics extra self belief in guiding scholar explorations and questions.

The textual content comprises 1000's 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 person has entry to the program).

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**Extra info for Advanced Euclidean Geometry**

**Example text**

In any case, the statements made in the proof hold for both diagrams. Ceva’s theorem is an equivalence (or biconditional) and therefore requires two proofs (one the converse of the other). We will first prove that if the three lines containing the vertices of AABC and intersecting the opposite sides at points L, , . , ^ AN BL CM ,, , M, and N, respectively^ are concurrent, then = 1. We offer three proofs. The first (though not the simplest) requires no auxiliary lines. Q ro o f I In Figure 2-4, AL, BM, and CN meet at point P.

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.

Construct perpendicular bisectors OP and OQ of sides DC and AB at points P and Q, respectively. Point N is on PO . ) Because O is a point on the perpendicular bisector of DC, DO = CO. Similarly, OA = OB. We began with AD = BC. Therefore AADO = ABCO (SSS) and mAAOD = mABOC. We can easily establish that mADOP = mACOP. By addition, mAAOP = mABOP. The supplements of these angles are also equal in measure: mAAON = mABON. But because AAOQ = ABOQ (SSS), mAAOQ = mABOQ. Because the angle bisector is unique, ON and OQ must coincide and the perpendiculars to these must also be parallel.