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Plane Geometry - Bruce E. Shapiro

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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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