Download Deductive Geometry

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
9 Deductive Geometry
Example 1T
In the figure, AOB, COD and EOF are straight lines.
AOC = 90. Prove that a + b = 90.
Solution:
AOE = BOF = b
AOC + AOE + DOE = 180
90 + b + a = 180
a + b = 90
vert. opp. s
adj. s on st. line
9.2 Deductive Proofs Related to Lines and Triangles
A. Angles Related to Intersecting Lines
9 Deductive Geometry
Example 2T
In the figure, AFB is a straight line. If CFB = AFD,
prove that CFD is a straight line.
Solution:
AFD + DFB = 180
CFB = AFD
Consider
CFB + DFB
 AFD + DFB
 180
∴ CFD is a straight line.
adj. s on st. line
given
given
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
9 Deductive Geometry
Example 3T
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
a = 30
b + 120 = 180
b = 60
Consider ACD = a + b
= 30 + 60
= 90
∴ AC  CD
int. s, AB // CF
int. s, CF // DE
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
9 Deductive Geometry
Example 4T
In the figure, AGHB, CGD and EHF are straight lines.
Prove that CD // EF.
Solution:
DGH  AGC
x
∵ DGH  FHB  x
∴ CD // EF
vert. opp. s
corr. s equal
9 Deductive Geometry
Example 5T
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
9 Deductive Geometry
Example 6T
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
40 + CBD  120
CBD  80
Consider ABC  CBD + ABD
 80 + 100
 180
∴ ABC is a straight line.
ext.  of 
adj. s supp.
9.3
Deductive Proofs Related to Congruent and
Isosceles Triangles
A. Congruent Triangles
9 Deductive Geometry
Example 7T
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
CB = CD
AC = AC
∴ ABC  ADC
(b) ∴ DCA = BCA
given
given
common side
SSS
corr. s,  s
9 Deductive Geometry
Example 8T
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
AB = CB
BAE = BCD
∴ ABE  CBD
(b) ∵ AB = CB
and BD = BE
∴ AB – BD = CB – BE
AD = CE
common 
given
given
ASA
given
corr. sides,  s
9 Deductive Geometry
Example 8T
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
DAF = ECF
AD  CE
∴ ADF  CEF
∴ DF = EF
vert. opp. s
given
proved
AAS
corr. sides,  s
9.3
Deductive Proofs Related to Congruent and
Isosceles Triangles
B. Isosceles Triangles
9 Deductive Geometry
Example 9T
In the figure, AB = AC and ACD = DCB.
Prove that ADC = 3ACD.
Solution:
In ABC,
∵ ACD = DCB
∴ ACB  2ACD
∴ ABC = ACB
= 2ACD
In BCD,
ADC = ABC + DCB
= 2ACD + ACD
= 3ACD
given
base s, isos. 
ext.  of 
9.3
Deductive Proofs Related to Congruent and
Isosceles Triangles
B. Isosceles Triangles
9 Deductive Geometry
Example 10T
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
BAD = BDA
BDC + BDA  90
BDC  90 – BDA
= 90 – BAD
In ACD, ACD + CAD + ADC = 180
BCD + BAD + 90 = 180
∴ BCD  90 – BAD
∵ BCD = BDC
∴ BC = BD
∴ BCD is an isosceles triangle.
given
base s, isos. 
given
 sum of 
proved
sides opp. eq. s
9 Deductive Geometry
Example 11T
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
BD = CD
ADB = ADC
 ADB  ADC
(b) In ABC,
∵ AB  AC
∴ ABC  ACB
common side
given
given
SAS
corr. sides,  s
base s, isos. 
9.4
Deductive Proofs Related to Similar Triangles
9 Deductive Geometry
Example 12T
In the figure, ABC and AED are straight lines.
AEB = ADC. Prove that ABE ~ ACD.
Solution:
∵ AEB = ADC
∴ BE // CD
In ABE and ACD,
BAE = CAD
ABE = ACD
AEB  ADC
∴ ABE ~ ACD
given
corr. s equal
common 
corr. s, BE // CD
given
AAA
9 Deductive Geometry
Example 13T
In the figure, ABD and CBE are straight lines.
Prove that ABE ~ DBC.
Solution:
In ABE and DBC,
AB 2 1
 
DB 6 3
BE 3 1
 
BC 9 3
ABE = DBC
∴ ABE ~ DBC
vert. opp. s
ratio of 2 sides, inc. 
9 Deductive Geometry
Example 14T
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,
AB 4 5


AE 3.2 4
BC 5

ED 4
AC 4  1 5


AD
4
4
AB BC AC 5
∵
=
=
=
AE ED AD 4
∴ ABC ~ AED
3 sides proportional
9 Deductive Geometry
Example 14T
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
ABC + BAC + ACB = 180
BAC + BAC + 47 = 180
BAC = 66.5
∴ EAD = BAC = 66.5
BAE = BAC + EAD
= 66.5 + 66.5
= 133
base s, isos. 
 sum of 
corr. s, ~ s