1. Theorem 1: A B C D a b
If a = b, then AB // CD (alt. s eq.)
Theorem 4 : A B C D a b
If AB // CD then a = b (alt. s, AB // CD)

2. Theorem 2: A B C D a b
If a = b, then AB // CD (corr. s eq.)
Theorem 5 : A B C D a b
If AB // CD then a = b (corr. s, AB // CD)

3. Theorem 3: A B C D a b
If a + b = 1800, then AB // CD (int. s supp.)
Theorem 6 : A B C D a b
If AB // CD then a + b = 1800 (int. s, AB // CD)

4. Theorem 7 :
If AB // CD and EF // CD then AB // EF (transitive property of // lines) A B E D C F a A B
Proof: a = b (corr. s, AB // CD) c = b (corr. s, EF // CD) i.e. a = c AB // EF (corr. s, eq) b  C D c E F 

5. Example 1: In the figure, AB // CD and b = d. Prove that CB // ED
Proof: Let BCD = c. AB // CD (given) b = c (alt. s, AB // CD) b = d (given) i.e. c = d CB // ED (alt. s, eq) 

6. Example 2:ABC =1800 ,BCD=700 and CDE=1600 . Prove that AB // ED
Proof: Construct FG passing through C and parallel to AB FG // AB (by construction) ABC +x=1800 (int. s, AB // FC) x=  x = x + y = (given)  y = 200  CDE+y = =1800  FG // ED (int. s, supp) FG // AB and FG // ED  AB // ED (transitive property of // lines) A B C E D
1300
700
1600 F G x y 

7. Example 3: In ABC, BD=DE and DE // BC
Example 3: In ABC, BD=DE and DE // BC. Prove the BE is the angle bisector of ABC .
Proof: DBE =DEB (base s, isos. ) DEB =EBC (alt s, DE//BC) DBE =EBC BE is the angle bisector of ABC .   A  D E B C

8. Theorem 8: B A C d a b
d = a + b (ext. s of ) D B A C D a b E
Proof: Construct CE //AB f g
g = b (corr, s , BA//CE ) f = a (alt, s , BA//CE ) 
f + g = a+b
d = a+b

9. Theorem 9: B A a b c
The sum of interior angles is (ext. s of ) A
Proof: Construct CE //AB a
d+c = 1800 (adj. s on a st. line ) d = a+b (ext. s of ) d b c B C D 
a + b + c = 1800

10. Example 4: AED and BEC are straight lines and x + y = 900
Example 4: AED and BEC are straight lines and x + y = 900. Prove the BE is the angle bisector of ABC .
In ABC, p+2x+y = 1800 ( sum of )  p = 1800 –2x-y ……(1) In ADC, x+2y+q = 1800 ( sum of )  q = 1800 –x-2y ……(1)
(1) + (2), p+q = (1800 –2x-y )+ (1800 –x-2y ) = 3600 –3(x+y ) = 3600 –3(900) (given) = 900 B A C D E x y p q  

11. Example 5: AEF, AEC and BED are straight lines and p + q = 1200
Example 5: AEF, AEC and BED are straight lines and p + q = Prove that y = x D A E C B F q y p x
In ABD,  DBF=p+2x (ext.  of )  2y = p+2x ……(1) In ABC,  CBF=q+x (ext.  of )  y = q+x ……(2)
(1) + (2), 2y+y = p+2x+q+x 3y = 3x +p+q 3y = 3x (given) y = x+400

12. Similar Triangles  ABC DEF  (corr. sides,  s)
If 2 triangles are similar then their corresponding sides same in ratio 
ABC DEF
(Given) 
(corr. sides,  s)
If the corresponding sides of two triangles are proportional, then these triangles are similar. (3 sides proportional)  
ABC DEF
(3 sides prop.)

13. If 2 triangles are similar then their corresponding angles are equal
ABC DEF
(Given) 
ABC=DEF
BCA=EFD
CAB=FDE
(corr.  s,  s)
If the corresponding angles are equal then these triangles are similar. 
ABC=DEF
BCA=EFD
CAB=FDE 
ABC DEF
(equiangular or AAA)

14. If 2 triangles are similar then two corresponding sides are same in ratio and included angle is equal 
ABC DEF
(Given) 
(corr. sides,  s)
ABC=DEF
(corr.  s,  s)
If two corresponding sides of two triangles are of same ratio and the included angles are equal, then these triangles are similar. 
ABC=DEF 
ABC DEF
(ratio of 2 sides, inc. )

15. A B C
4.5 6
7.5
370
530
Example 6 3 P Q R 4 5
(a) Prove that ABC PQR
(b) Find P
Solution (a)
Solution (b) 
A =1800 ( sum of )
A=900 
P= A
(corr.  s,  s)
P= 900 
i.e. 
ABC PQR
(3 sides prop.)

16. A B C H K 4 6 a b c h k 3
Example 7
In ABC, HK//BC, HK=4, BC=6 and HB=3.
(a) Prove that AHK  ABC
(b) Find that length of AH
Solution (b)
Solution (a)
(corr. sides,  s)
In AHK and ABC 
k=c
(corr.  s, HK//BC)
h=b
(corr.  s, HK//BC)
a=a
(common angle)
6AH=12+4AH 
AHK  ABC
(A.A.A.)
2AH=12 
AH=6

17. Example 8 D A B C 15 9 6 K 12 8
AKC and DKB are straight lines
(a) Prove that ABK  CDK
(b) Find that length of AB
Solution (b)
Solution (a)
(corr. sides,  s) 
i.e.
(vert. opp,  s)
AKB=CKD 
ABK  CDK
(ratio of 2 sides, inc. )

18. CH 10B (Q10) A B D F E C 22 14 12 p q s r 8 r = 11 (corr. sides,  s)
CBD=EBF
(common angle)
BCD=BEF
(corr.  s, CD//EF)
BDC=BFE
(corr.  s, CD//EF) 
BCD  BEF
(A.A.A)
(corr. sides,  s)
In FCD and FAB
CFD=AFB
(common angle)
FCD=FAB
(corr.  s, CD//AB)
FDC=FBA
(corr.  s, CD//AB) 
FCD  FAB
(A.A.A)
(corr. sides,  s)
(corr. sides,  s)
(corr. sides,  s)
r = 11