Plane Geometry - Bruce E. Shapiro

SECTION 33. RECTANGLES 16933.1 and drop the perpendicular from C to D as shown. Then triangles△ADC and △BDC are right.By additivity of defect (30.11),0 = δ(△ABC) = δ(△ADC) + δ(△BDC)Since defect is non-negative, each of the two sub-triangles also have zerodefect. Hence there is a right triangle with zero defect (statement (2)).[(2) ⇒ (3)] Let △ABC be a right triangle with zero defect (hypothesis,statement (2)).Designate B as the right angle. Then∠BAC + ∠ACB = 90Since it is possible to construct a congruent copy of any triangle on eitherside of a congruent base (see theorem 23.3), there exists a point D on theopposite side of ←→ AC from B such that△ABC ∼ = △ADCFigure 33.2: A rectangle constructed from two right triangles in proof of(2) ⇒ (3).(See figure 33.2.) By congruency,α = ɛγ = ζα + γ = 90 (Since δ(△ABC) = 0)ɛ + ζ = 90 (substitution)α + ζ = 90ɛ + γ = 90The last two lines complete the proof that □ABCD is a rectangle. Hencea rectangle exists (statement (3)).Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.