9 Deductive Geometry 9.1 Introduction to Deductive Reasoning and Proofs 9.2 Deductive Proofs Related to Lines and Triangles 9.3 Deductive Proofs Related to Congruent and Isosceles Triangles 9.4 Deductive Proofs Related to Similar Triangles

9.1 Introduction to Deductive Reasoning and Proofs A. What is Deductive Reasoning?

9.1 Introduction to Deductive Reasoning and Proofs A. What is Deductive Reasoning?

9.1 Introduction to Deductive Reasoning and Proofs B. Euclid and ‘Elements’

9.1 Introduction to Deductive Reasoning and Proofs B. Euclid and ‘Elements’

9.1 Introduction to Deductive Reasoning and Proofs C. Deductive Proofs of Theorems

9.1 Introduction to Deductive Reasoning and Proofs C. Deductive Proofs of Theorems

9.2 Deductive Proofs Related to Lines and Triangles A. Angles Related to Intersecting Lines

9.2 Deductive Proofs Related to Lines and Triangles A. Angles Related to Intersecting Lines

9.2 Deductive Proofs Related to Lines and Triangles A. Angles Related to Intersecting Lines

Example 1T Solution: 9 Deductive Geometry In the figure, AOB, COD and EOF are straight lines. AOC = 90. Prove that a + b = 90. Solution: AOE = BOF = b vert. opp. s AOC + AOE + DOE = 180 adj. s on st. line 90 + b + a = 180 a + b = 90

9.2 Deductive Proofs Related to Lines and Triangles A. Angles Related to Intersecting Lines

Example 2T Solution: 9 Deductive Geometry In the figure, AFB is a straight line. If CFB = AFD, prove that CFD is a straight line. Solution: AFD + DFB = 180 adj. s on st. line CFB = AFD given Consider CFB + DFB AFD + DFB given 180 ∴ CFD is a straight line. adj. s supp.

9.2 Deductive Proofs Related to Lines and Triangles B. Angles Related to Parallel Lines

9.2 Deductive Proofs Related to Lines and Triangles B. Angles Related to Parallel Lines

9.2 Deductive Proofs Related to Lines and Triangles B. Angles Related to Parallel Lines

Example 3T Solution: 9 Deductive Geometry In the figure, AB // DE, BAC = 150 and EDC = 120. Prove that AC  CD. Solution: Construct a line CF such that AB // CF // DE. a + 150 = 180 int. s, AB // CF a = 30 b + 120 = 180 int. s, CF // DE b = 60 Consider ACD = a + b = 30 + 60 = 90 ∴ AC  CD

9.2 Deductive Proofs Related to Lines and Triangles B. Angles Related to Parallel Lines

9.2 Deductive Proofs Related to Lines and Triangles B. Angles Related to Parallel Lines

9.2 Deductive Proofs Related to Lines and Triangles B. Angles Related to Parallel Lines

Example 4T Solution: 9 Deductive Geometry In the figure, AGHB, CGD and EHF are straight lines. Prove that CD // EF. Solution: DGH  AGC vert. opp. s  x ∵ DGH  FHB  x ∴ CD // EF corr. s equal

Example 5T Solution: 9 Deductive Geometry In the figure, BAC = 25, reflex ACD = 305 and CDE = 30. Prove that AB // DE. Solution: Construct a line FC such that AB // FC. a  BAC alt. s, AB // FC  25 b  360 – 305 – a s at a pt.  360 – 305 – 25  30 ∵ CDE  b  30 ∴ FC // DE alt. s equal ∴ AB // DE

9.2 Deductive Proofs Related to Lines and Triangles C. Angles Related to Triangles

9.2 Deductive Proofs Related to Lines and Triangles C. Angles Related to Triangles

Example 6T Solution: 9 Deductive Geometry In the figure, EDB is a straight line. ABD = 100, EDC = 120 and DCB = 40. Prove that ABC is a straight line. Solution: In BCD, BCD + CBD  CDE ext.  of  40 + CBD  120 CBD  80 Consider ABC  CBD + ABD  80 + 100  180 ∴ ABC is a straight line. adj. s supp.

9.3 Deductive Proofs Related to Congruent and Isosceles Triangles A. Congruent Triangles

Example 7T Solution: 9 Deductive Geometry In the figure, AD = AB and CD = CB. (a) Prove that ABC  ADC. (b) Prove that DCA = BCA. Solution: (a) In ABC and ADC, AB = AD given CB = CD given AC = AC common side ∴ ABC  ADC SSS (b) ∴ DCA = BCA corr. s,  s

Example 8T Solution: 9 Deductive Geometry In the figure, BAE = BCD and AB = BC.   (a) Prove that ABE  CBD.   (b) Prove that DF = EF. Solution: (a) In ABE and CBD, ABE = CBD common  AB = CB given BAE = BCD given ∴ ABE  CBD ASA (b) ∵ AB = CB given and BD = BE corr. sides,  s ∴ AB – BD = CB – BE AD = CE

Example 8T Solution: 9 Deductive Geometry In the figure, BAE = BCD and AB = BC.   (a) Prove that ABE  CBD.   (b) Prove that DF = EF. Solution: In ADF and CEF, AFD = CFE vert. opp. s DAF = ECF given AD  CE proved ∴ ADF  CEF AAS ∴ DF = EF corr. sides,  s

9.3 Deductive Proofs Related to Congruent and Isosceles Triangles B. Isosceles Triangles

Example 9T Solution: 9 Deductive Geometry In the figure, AB = AC and ACD = DCB. Prove that ADC = 3ACD.   Solution: In ABC, ∵ ACD = DCB given ∴ ACB  2ACD ∴ ABC = ACB base s, isos.  = 2ACD In BCD, ADC = ABC + DCB ext.  of  = 2ACD + ACD = 3ACD

9.3 Deductive Proofs Related to Congruent and Isosceles Triangles B. Isosceles Triangles

Example 10T Solution: 9 Deductive Geometry In the figure, ABC is a straight line. AB = DB and ADC = 90. Prove that BCD is an isosceles triangle. Solution: In ABD, ∵ AB = DB given ∴ BAD = BDA base s, isos.  ∵ BDC + BDA  90 given ∴ BDC  90 – BDA = 90 – BAD In ACD, ACD + CAD + ADC = 180  sum of  BCD + BAD + 90 = 180 ∴ BCD  90 – BAD ∵ BCD = BDC proved ∴ BC = BD sides opp. eq. s ∴ BCD is an isosceles triangle.

Example 11T Solution: 9 Deductive Geometry In the figure, BD = CD and ADB = ADC. (a) Prove that ADB  ADC. (b) Prove that ABC = ACB. Solution: (a) In ADB and ADC, AD = AD common side BD = CD given ADB = ADC given  ADB  ADC SAS (b) In ABC, ∵ AB  AC corr. sides,  s ∴ ABC  ACB base s, isos. 

9.4 Deductive Proofs Related to Similar Triangles

Example 12T Solution: 9 Deductive Geometry In the figure, ABC and AED are straight lines. AEB = ADC. Prove that ABE ~ ACD. Solution: ∵ AEB = ADC given ∴ BE // CD corr. s equal In ABE and ACD, BAE = CAD common  ABE = ACD corr. s, BE // CD AEB  ADC given ∴ ABE ~ ACD AAA

Example 13T Solution: 9 Deductive Geometry In the figure, ABD and CBE are straight lines. Prove that ABE ~ DBC. Solution: In ABE and DBC, ABE = DBC vert. opp. s ∴ ABE ~ DBC ratio of 2 sides, inc. 

Example 14T Solution: 9 Deductive Geometry In the figure, ACB = 47. AB = 4, BC = 5, CD = 1, AD = DE = 4 and AE = 3.2. Find BAE. Solution: In ABC and AED, ∵ ∴ ABC ~ AED 3 sides proportional

Example 14T Solution: 9 Deductive Geometry In the figure, ACB = 47. AB = 4, BC = 5, CD = 1, AD = DE = 4 and AE = 3.2. Find BAE. Solution: In ABC, ∵ AC = 4 + 1 = 5 = BC ∴ ABC = BAC base s, isos.  ABC + BAC + ACB = 180  sum of  BAC + BAC + 47 = 180 BAC = 66.5 ∴ EAD = BAC = 66.5 corr. s, ~ s BAE = BAC + EAD = 66.5 + 66.5 = 133