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By Alfred S. Posamentier

Advanced Euclidean Geometry provides a radical evaluate of the necessities of high institution geometry  after which expands these strategies to complicated Euclidean geometry, to offer academics extra self belief in guiding pupil explorations and questions.

The textual content includes hundreds and hundreds of illustrations created within the Geometer's Sketchpad Dynamic Geometry® software program. it's 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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Parallelograms ABGF and ACDE are constructed on sides AB and AC of AABC (see Figure 1-34). ) DE and GE intersect at point P. ^ in g BC as a side, construct parallelogram BCJK so that BiC 11PA and BK = PA. From this configuration. d . 300) proposed an extension of the Pythagorean theorem. He proved that the sum of the area of parallelogram ABGF and the area of parallelogram ACDE is equal to the area of parallelogram BCJK. Prove this relationship. ) FIGURE 1-34 Chapter 1 ELEMENTARY EUCLIDEAN GEOMETRY REVISITED 23 4.

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.

The following application demonstrates a somewhat different use of Ceva s theorem. A pplication 5 Q roof In AABCy where CD is the altitude to AB and P is any point on DC, AP intersects CB at point Q and BP intersects CA at point R (see Figure 2-11). Prove that ARDC = AQDC. • Let DR and DQ intersect the line containing C and parallel to AB, at points G and H, respectively. ACGR ~ AADR ^ RA AD BQ DB ASD Q ~A CH Q =>^ = - H (I) (II) 36 ADVANCED EUCLIDEAN GEOMETRY We now apply Ceva’s theorem to AABC to get: CR AD BQ = 1 Ra ' Db ' QC (III) Substituting (I) and (II) into (III) gives us: ^ ad ^ DB_ ’ db ' CH ~ ^ GC _ CH “ ^ This implies that GC = CH.

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