Plane Geometry - Bruce E. Shapiro
Plane Geometry - Bruce E. Shapiro Plane Geometry - Bruce E. Shapiro
170 SECTION 33. RECTANGLESFigure 33.3: Construct an arbitrariy large rectangle. Start with a givenrectangle, such as the one in the lower left corner, and then constructlarger rectangles by building congruent triangles on common edges. Therepeat the process, doubling the size of the rectangle at each step, until therequired size is reached.[(3) ⇒ (4)] (sketch of proof) By an arbitrarily large rectangle we mean thatfor any number M there is a rectangle all of whose sides are larger than M.From (3) there exists a rectangle. We can split this into two right triangles,and then proceed to duplicate the rectangle on one of the original rectangle’sedge by constructing congruent right triangles. This creates a rectangle oftwice the width of the first triangle (see figure 33.3).Then keep repeating the process until we have an edge longer than M(Archimedian property).Then repeat the process on the other direction, so that both dimensions arelarger than M. Thus by construction, an arbitrarily large triangle exists(statement (4)).[(4) ⇒ (5)] Let △ABC be a right triangle with right angle at C. By (4)there exists a rectangle □CDEF such that DC > AC and F C > BC.Since □CDEF is a parallelogram, it is convex (theorem 31.9).Since it is a rectangle, all four angles are 90 (definition of rectangle) andhence δ(□ABCD) = 0.By the additivity of defect (theorem 31.6),δ(△DEF ) = δ(△DCF ) = 0Similarly0 = δ(△CDF ) = δ(△ADF ) + δ(△ACF )« CC BY-NC-ND 3.0. Revised: 18 Nov 2012
SECTION 33. RECTANGLES 171Figure 33.4: Every right triangle has zero defect.hence δ(△ACF ) = 0. Applying additivity of defect one more time,0 = δ(△ACF ) = δ(△ABF ) + δ(△ABC)hence δ(△ABC) = 0. Since △ABC was an arbitrary right triangle, we canconclude that every right triangle has zero defect (statement (5)).[(5) ⇒ (6)] By the lemma we can divide any triangle into right triangles.By (5) every right triangle has zero defect. By additivity of defect, everytriangle has zero defect (statement (6)).Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.
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170 SECTION 33. RECTANGLESFigure 33.3: Construct an arbitrariy large rectangle. Start with a givenrectangle, such as the one in the lower left corner, and then constructlarger rectangles by building congruent triangles on common edges. Therepeat the process, doubling the size of the rectangle at each step, until therequired size is reached.[(3) ⇒ (4)] (sketch of proof) By an arbitrarily large rectangle we mean thatfor any number M there is a rectangle all of whose sides are larger than M.From (3) there exists a rectangle. We can split this into two right triangles,and then proceed to duplicate the rectangle on one of the original rectangle’sedge by constructing congruent right triangles. This creates a rectangle oftwice the width of the first triangle (see figure 33.3).Then keep repeating the process until we have an edge longer than M(Archimedian property).Then repeat the process on the other direction, so that both dimensions arelarger than M. Thus by construction, an arbitrarily large triangle exists(statement (4)).[(4) ⇒ (5)] Let △ABC be a right triangle with right angle at C. By (4)there exists a rectangle □CDEF such that DC > AC and F C > BC.Since □CDEF is a parallelogram, it is convex (theorem 31.9).Since it is a rectangle, all four angles are 90 (definition of rectangle) andhence δ(□ABCD) = 0.By the additivity of defect (theorem 31.6),δ(△DEF ) = δ(△DCF ) = 0Similarly0 = δ(△CDF ) = δ(△ADF ) + δ(△ACF )« CC BY-NC-ND 3.0. Revised: 18 Nov 2012