Plane Geometry - Bruce E. Shapiro
Plane Geometry - Bruce E. Shapiro Plane Geometry - Bruce E. Shapiro
116 SECTION 24. EXTERIOR ANGLESTheorem 24.4 (Exterior Angle Theorem a ) The measure of any exteriorangle of a triangle is strictly greater than the measure of either of itsremote interior angles.a Euclid’s Proposition 16In figure 24.1 this means that ɛ > β, ɛ > γ, δ > β, δ > γ.Figure 24.2: Definitions for proof of theorem 24.4Proof. See figure 24.2. We will show that α > β. The proof that α > γ isanalogous.Let E ∈ AC be its midpoint (Every segment has a midpoint).Choose F ∈ −→ BE such that BE = EF . This is possible by the ruler postulate.By the vertical angle theorem δ = ɛ.By SAS △BEA ∼ = △F EC (because BE = EF, δ = ɛ, and AE = EC), andso ∠ζ = ∠β.Since F and B are on opposite sides of ←→ AC (because F ∗ E ∗ B) and B andD are also on opposite sides of line ←→ AC (since B ∗ C ∗ D), then D and F areon the same side of ←→ AC by the plane separation postulate, i.e, F ∈ H ←→ D, ACSimilarly, since A and E are on the same side ←→ CD (since A ∗ E ∗ C), andE and F are on the same side of ←→ CD (since B ∗ E ∗ F ), then A and F areon the same side of the line ←→ CD, also by the plane separation postulate.Hence F ∈ H A,←→ CD.Hence F ∈ H D,←→ AC∪ H A,←→ CD.By the definition of the interior of an angle, this means F is in the interiorof ∠α. Hence ζ < α by the betweenness theorem for rays. Since ζ = β wehave β < α which is what the theorem states.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012
SECTION 24. EXTERIOR ANGLES 117Theorem 24.5 (Uniqueness of Perpendiculars) For every line l andfor every point P ∉ l, there exists exactly one line m ⊥ l such that P ∈ m.Proof. See figure 24.3. Suppose P is a point with P ∉ l and let m ⊥ lthrough P .Suppose n ⊥ l is a second line, distinct from m, through P . (RAA assumption)Let R be the point at which n intersects l, Then R ≠ Q. (Otherwise,R ∈ m and there would be two distinct points, R and P , at which m andn intersect. This would mean that m = n which would violate the RAAassumption.)Triangle △P RQ has an exterior angle β = 90 and an interior angle α = 90at R by the RAA assumption. This contradicts the exterior angle theorem(theorem 24.4) which says that β > α. Hence we reject the RAA hypothesis.Figure 24.3: Since there can be only one perpendicular line to l throughthe point P , β ≠ 90.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.
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116 SECTION 24. EXTERIOR ANGLESTheorem 24.4 (Exterior Angle Theorem a ) The measure of any exteriorangle of a triangle is strictly greater than the measure of either of itsremote interior angles.a Euclid’s Proposition 16In figure 24.1 this means that ɛ > β, ɛ > γ, δ > β, δ > γ.Figure 24.2: Definitions for proof of theorem 24.4Proof. See figure 24.2. We will show that α > β. The proof that α > γ isanalogous.Let E ∈ AC be its midpoint (Every segment has a midpoint).Choose F ∈ −→ BE such that BE = EF . This is possible by the ruler postulate.By the vertical angle theorem δ = ɛ.By SAS △BEA ∼ = △F EC (because BE = EF, δ = ɛ, and AE = EC), andso ∠ζ = ∠β.Since F and B are on opposite sides of ←→ AC (because F ∗ E ∗ B) and B andD are also on opposite sides of line ←→ AC (since B ∗ C ∗ D), then D and F areon the same side of ←→ AC by the plane separation postulate, i.e, F ∈ H ←→ D, ACSimilarly, since A and E are on the same side ←→ CD (since A ∗ E ∗ C), andE and F are on the same side of ←→ CD (since B ∗ E ∗ F ), then A and F areon the same side of the line ←→ CD, also by the plane separation postulate.Hence F ∈ H A,←→ CD.Hence F ∈ H D,←→ AC∪ H A,←→ CD.By the definition of the interior of an angle, this means F is in the interiorof ∠α. Hence ζ < α by the betweenness theorem for rays. Since ζ = β wehave β < α which is what the theorem states.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012