Criteria for Similarity of Triangles

MATHS TRIANGLES CRITERIA FOR SIMILARITY OF TRIANGLES CRITERIA FOR SIMILARITY OF TRIANGLES (i) (AAA Similarity) If two triangles are equiangular, then they are similar. (ii) (SSS Similarity) If the corresponding sides of two triangles are proportional, then they are similar. (iii) (SAS Similarity) If in two triangle’s one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar. ANGLE-ANGLE-ANGLE SIMILARITY (AAA-SIMILARITY) (THEOREM) : Statement : In two triangles, if the corresponding angles are equal then their corresponding sides are in the same ratio (or proportion) and hence two triangles are similar. OR, Two equiangular triangles are similar. D A B Given : C E F DABC and DDEF are equiangular. Hence, A = D, B = E and C = F To prove : ABC ~ DEF Proof : Here, ABC and DEF are equiangular, i.e., A = D, B = E and C = F .....(1) Three cases arises for sides AB of ABC and DE of DEF 1 MATHS (i) AB = DE (ii) Case (1) : When AB = DE Proof : In ABC and DEF A and D (Given) AB = DE (Given) B and E (Given) AB > DE (iii) AB < DE Then by ASA rule of congruence, ABC DEF Therefore BC = EF, AC = DE, AB = DE AB = BC = CA DE EF DF ABC ~ DEF Case (2) : When AB > DE Construction : As in figure taking the point P and Q on side AB and AC such that AP = DE and AQ = DF. D A P Q B Proof : C E F In DAPQ and DDEF AP = DE (By Construction) AQ = DF (By Construction) A = D (Given) Therefore by Side-Angle-Side Rule for congruency APQ DEF So, APQ = E But B and E  APQ = B, which is corresponding angle .....(1) (Given) .....(2) Consequently, PQ || BC Hence AP = AQ AB AC (By Basic Proportionality Theorem) 2 MATHS AP = AB AQ AC AP = AQ Also, DE DF AP = DE  AQ DF .....(3) (By Construction) .....(4) From (3) and (4), AB = DE AC DF AB = AC  DE DF AB = BC Similarly, DE EF  .....(5) .....(6) From (5) and (6), we get AB AC BC = = DE DF EF Hence DABC ~ DDEF Case (3) : When AB < DE, proof is the same as for case (2). COROLLARY : ANGLE-ANGLE SIMILARITY (AA SIMILARITY) If two angles of one triangle are respectively equal to two angles of another triangle than the two triangles are similar. Ex.1 In a trapezium ABCD AB||DC and DC = 2AB. EF drawn parallel to AB cuts AD in F and BC in E such that Sol. BE= 3 . Diagonal DB intersects EF at G. Prove that 7FE = 10AB. EC 4 In DFG and DAB, 1 = 2 [Corresponding s  AB || FG] FDG = ADB [Common] 3 MATHS  DFG ~ DAB [By AA rule of similarity]  DF= FG DA AB .....(i) Again in trapezium ABCD EF||AB||DC  AF= BE DF EC  AF= 3 DF 4  AF=1= 3 +1 DF 4  AF+DF= 7 DF 4  AD= 7 DF 4  DF= 4 AD 7 BE= 3 (given )  EC 4  …….(ii) From (i) and (ii), we get FG= 4 i.e. FG = 4 AB 7 AB 7 ......(iii) In BEG and BCD, we have BEG = BCD [Corresponding angle  EG||CD] GBE = DBC [Common]  BEG ~ BCD [By AA rule of similarity]  BE= EG BC CD  3 = EG 7 CD  BE= 3 i.e..EC= 4 EC+BE= 4+3 BC= 7  EG 7 BE 3 BE 3  BE 3 4 MATHS  EG= 3 CD= 3(2AB) CD=2AB(given ) 7 7  EG= 6 AB 7 .....(iv) Adding (iii) and (iv), we get FG+EG= 4 AB+ 6 AB= 10AB 7 7 7  EF= 10ABi.e., 7EF=10AB. 7 Hence proved. SIDE-SIDE-SIDE SIMILARITY (SSS-SIMILARITY) (THEOREM) : Statement : If the corresponding sides of two triangles are proportional (i.e. in the same ratio), their corresponding angles are equal and hence the two triangles are similar. Given : ABC and DEF AB = BC = AC 1 DE EF DF D A P B To prove : C Q E F ABC ~ DEF Construction : Taking points P on DE and Q on DF such that DP = AB and DQ = AC, then join PQ. Proof : In ABC ~ DEF AB = AC = BC DE DF EF .....(1) 5 MATHS AB DE = AC DF .....(2) AB AC = DP DQ (By Construction) AB DP = AC DQ .....(3) From (2) and (3), DE DP = DF DQ DP DQ = DE DF Therefore, by basic proportionality theorem, PQ || EF So, DPQ = DEF and DQP = DFE Hence by AA similarity, DPQ ~ DEF (Corresponding angles) .....(4) Hence the corresponding sides of similar triangles DDPQ and DDEF are proportional. i.e. DP PQ = DE EF AB PQ = DE EF .....(5) From (1) and (5), PQ BC = EF EF PQ = BC ......(6) Now, in ABC and DPQ AB = DP (By Construction) AC = DQ (By Construction) BC = PQ [From (6)] So by SSS congruence rule 6 MATHS ABC DPQ .....(7) From (4) and (7) ABC ~ DPQ ~ DEF ABC ~ DEF SIDE-ANGLE-SIDE SIMILARITY (SAS-SIMILARITY) (THEOREM) : Statement : If one angle of one triangle is equal to an angle of other triangle and if the side including the angles are proportional, then the two triangles are similar. Given : ABC and DEF, such that A = D AB AC = 1 and DE DF To prove : ABC ~ DEF Construction : Taking points P and Q on sides DE and DF respectively such that AB = DP andAC=DQ, join PQ. A D P B Proof : C Q E F In ABC and DPQ AB = DP (By Construction) AC = DQ (By Construction) A = D (Given) By SAS rule of congruence ABC DPQ AB = AC DP DQ AB = AC and DE DF .....(1) (Given) .....(2) (By Construction) .....(3) 7 MATHS From (2) and (3), DP DE = DQ DF DP DQ = DE DF By converse of basic Proportionality theorem PQ || EF So, DPQ = E and DQP = F (Corresponding angles) Consequently, by AA similarity, DPQ ~ DEF .....(4) From (1) and (4), we ABC ~ DPQ ~ DEF ABC ~ DEF Ex. 2 Prove that the ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides. A b c B Sol. P a q r CQ p R Given:  ABC and  PQR BC = a, CA = b, AB = c and QR = p, RP = q, PQ = r Also  ABC ~  PQR To prove : a = b = c = a+b+c p q r p+q+r Proof: Since  ABC and  PQR are similar, there for their corresponding sides are proportional a = b = c = k (say) p q r .....(i) 8 MATHS a = kp, b = kq, c = kr perimeter of ABC = a+b+c = kp+kq+kr perimeter of PQR p+q+r p+q+r k( p+q+r) = ..... (ii) ( p+q+r) =k From (i) and (ii) we get a = b = c = a+b+c = perimeter of ABC p q r p+q+r perimeter of PQR Ex.3 BAC = 900, AD is its bisector. IF DE ⊥ AC, prove that DE × (AB + AB) = AB × AC. Sol. It is given that AD is the bisector of A of ABC.  AB= BD AC DC  AB+1= BD+1 AC DC  AB+AC= BD+DC AC DC  AB+AC= BC AC DC [Adding 1 on both sides] ....(i) In ’s CDE and CBA, we have DCE = BCA [Common] DEC = BAC [Each equal to 900] So, by AA-criterion of similarity  CDE ~  CBA  CD= DE CB BA  AB= BC DE DC ....(ii) 9 MATHS From (i) and (ii), we have  AB+AC= AB AC DE  DE × (AB + AC) = AB × AC. 10