1. 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

2. 9.1 Introduction to Deductive Reasoning and Proofs
A. What is Deductive Reasoning?

3. 9.1 Introduction to Deductive Reasoning and Proofs
A. What is Deductive Reasoning?

4. 9.1 Introduction to Deductive Reasoning and Proofs
B. Euclid and ‘Elements’

5. 9.1 Introduction to Deductive Reasoning and Proofs
B. Euclid and ‘Elements’

6. 9.1 Introduction to Deductive Reasoning and Proofs
C. Deductive Proofs of Theorems

7. 9.1 Introduction to Deductive Reasoning and Proofs
C. Deductive Proofs of Theorems

8. 9.2 Deductive Proofs Related to Lines and Triangles
A. Angles Related to Intersecting Lines

9. 9.2 Deductive Proofs Related to Lines and Triangles
A. Angles Related to Intersecting Lines

10. 9.2 Deductive Proofs Related to Lines and Triangles
A. Angles Related to Intersecting Lines

11. 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

12. 9.2 Deductive Proofs Related to Lines and Triangles
A. Angles Related to Intersecting Lines

13. 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.

14. 9.2 Deductive Proofs Related to Lines and Triangles
B. Angles Related to Parallel Lines

15. 9.2 Deductive Proofs Related to Lines and Triangles
B. Angles Related to Parallel Lines

16. 9.2 Deductive Proofs Related to Lines and Triangles
B. Angles Related to Parallel Lines

17. 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

18. 9.2 Deductive Proofs Related to Lines and Triangles
B. Angles Related to Parallel Lines

19. 9.2 Deductive Proofs Related to Lines and Triangles
B. Angles Related to Parallel Lines

20. 9.2 Deductive Proofs Related to Lines and Triangles
B. Angles Related to Parallel Lines

21. 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

22. 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

23. 9.2 Deductive Proofs Related to Lines and Triangles
C. Angles Related to Triangles

24. 9.2 Deductive Proofs Related to Lines and Triangles
C. Angles Related to Triangles

25. 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.

26. 9.3 Deductive Proofs Related to Congruent and Isosceles Triangles
A. Congruent Triangles

27. 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

28. 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

29. 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

30. 9.3 Deductive Proofs Related to Congruent and Isosceles Triangles
B. Isosceles Triangles

31. 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

32. 9.3 Deductive Proofs Related to Congruent and Isosceles Triangles
B. Isosceles Triangles

33. 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.

34. 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. 

35. 9.4 Deductive Proofs Related to Similar Triangles

36. 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

37. 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. 

38. 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 sides proportional

39. 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 = = 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 
= 133