By Alfred S. Posamentier

*Advanced Euclidean Geometry* provides an intensive evaluate of the necessities of high university geometry after which expands these ideas to complex Euclidean geometry, to provide academics extra self assurance in guiding pupil explorations and questions.

The textual content comprises enormous quantities 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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**Extra info for Advanced Euclidean Geometry**

**Sample text**

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. , from point A): area AABL area A ACL Ж LC (I) area APBL area APCL Ж LC (II) Similarly: From (I) and (II): area AABL _ area APBL area AACL area APCL A basic property of proportions w y \X Z w —y \ ------- I provides that: OC 2/ BL _ area AABL - area APBL _ area AABP LC area AACL - area APCL area AACP We now repeat the process, using BM instead of AL: CM MA area ABMC area ABMA area APMC area АРМА (III) Chapter 2 CONCURRENCY of LINES in a TRIANGLE It follows that: CM MA area ABMC — area APMC area ABMA — area АРМА area ABCP area ABAP (IV) Once again we repeat the process, this time using CN instead of AL: AN NB area AACN area ABCN area AAPN area ABPN This gives us: AN NB area AACN — area AAPN area ABCN — area ABPN area AACP area ABCP (V) We now simply multiply (III), (IV), and (V) to get the desired result: BL LC CM AN _ area AABP area ABCP area AACP _ ^ ф MA NB area AACP area ABAP area ABCP By introducing an auxiliary line, we can produce a simpler proof.

1 (Menelaus^S theorem) The threepoints P, Q, 1 - ЛГАЛ 11. TlQ and BC, respectively, of /лАВС are collinear if andonly i f -----• ^ ^ ^ QB ^^ and R onthesidesAC,AB, BR CP — • — = —1. RC PA Like Ceva’s theorem, Menelaus’s theorem is an equivalence and therefore requires proofs for each of the two statements (converses of each other) that comprise the entire^theorem. We will first prove that if the three points P, Q, and R on the sides AC, AB, and BC,, respectively, of AABC are collinear, then AQ BR CP TT;: * ~ L We offer two proofs of this part of Menelaus’s theorem.

Because the angle bisector is unique, ON and OQ must coincide and the perpendiculars to these must also be parallel. Hence AB 11 CD. • Repeat the “proof” for O outside the quadrilateral. Then repeat the proof for O on DC. 22 ADVANCED EUCLIDEAN GEOMETRY 2. Discover the fallacy in the following “proof”: 45° = 60°. “P r o o f ” Construct equilateral triangle ABC. (See Figure 1-33). On side AB construct isosceles right triangle ADB with AB as hypotenuse. 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.