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)

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)

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)

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 

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) 

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) 1300 +x=1800  x = 500 x + y = 700 (given)  y = 200  CDE+y =1600+200 =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 

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

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

Theorem 9: B A a b c The sum of interior angles is 1800. (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

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  

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 = 1200. Prove that y = x + 400. 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+1200 (given) y = x+400

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.)

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)

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. )

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+530+370=1800 ( sum of ) A=900  P= A (corr.  s,  s) P= 900  i.e.  ABC PQR (3 sides prop.)

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

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. )

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