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Download Advanced Euclidean Geometry by Alfred S. Posamentier PDF

By Alfred S. Posamentier

Advanced Euclidean Geometry provides an intensive assessment of the necessities of high college geometry  after which expands these suggestions to complicated Euclidean geometry, to provide lecturers extra self belief in guiding scholar explorations and questions.

The textual content includes 1000s 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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Sample text

But mADBG = 45°, FIGURE 1-33 while mACBG = 60°; thus 45° = 60°. • 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.

P r o o f ” In quadrilateral ABCD, AD = BC. 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 bisector of an angle (interior or exterior) of a triangle partitions the opposite side proportionally to the two remaining sides, we have: AL ~ AB B M ~ BC CN ~ AC N Therefore: CL AM BN — • —— • —— AL BM CN BC AC AB . ---------- - — = —1 (because all three ratios are negative) AB BC AC ^ Thus, by Menelaus’s theorem, points I, M, and N are collinear. • A pplication 3 A circle through vertices B and C of AABC meets AB at point P and AC at <-> . , QC {RC){AC) point R. If PR meets BC at point Q, prove that = {PB){AB) A ro o f Consider AABC with transversal QPR (Figure 3-7).

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