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# Criteria for Similarity of Triangles

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#### Criteria for Similarity of Triangles - Lesson Summary

In general, there are several objects which have something common between them. Observing them closely, see that some of them have same shape but may have different or same size such figures are called similar figures.

In case of triangles “Two triangles are said to be similar if their corresponding angles are equal and corresponding sides are proportional”.

By using AAA similarity theorem, SSS similarity theorem and SAS similarity theorem we can prove two triangles are similar.

AAA similarity theorem or criterion:
If the corresponding angles of two triangles are equal, then their corresponding sides are proportional and the triangles are similar.

In ΔABC and ΔPQR, ∠A = ∠P , ∠B = ∠Q , and ∠C = ∠R  then  AB PQ =  BC QR =  ACPR and ΔABC ∼ ΔPQR.

Given: In ΔABC and ΔPQR, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R.

To prove:   AB PQ =  BC QR =  ACPR

Construction : Draw LM such that    PL AB =  PM AC .

Proof: In ΔABC and ΔPLM,

AB = PL and AC = PM (By Contruction)

∠BAC = ∠LPM (Given)

∴ ΔABC ≅ ΔPLM (SAS congruence rule)

∠B = ∠L (Corresponding angles of congruent triangles)

Hence ∠B = ∠Q (Given)

∴ ∠L =  ∠Q

LQ is a transversal to LM and QR.

Hence  ∠L =  ∠Q (Proved)

∴ LM ∥ QR

PL LQ =  PM MR

LQ PL =  MR PM   (Taking reciprocals)

LQ PL + 1 =  MR PM + 1  (Adding 1 to both sides)

LQ+PL PL =  MR+PM PM

PQ PL =  PR PM

PQ AB =  PR AC   (AB = PL and AC =PM)

AB PQ =  AC PR  (Taking Reciprocals) ............... (1)

AB PQ =  BC QR

AB PQ =  AC PR =   BC QR

∴ ΔABC ~ ΔPQR

SSS similarity theorem or criteria:
If sides of one triangle are proportional to the sides of the other triangles, then their corresponding angles are equal and the triangles are similar.

In ΔABC and ΔDEF,  AB DE =  BC EF =  AC DF then ∠A = ∠D , ∠B = ∠E , and ∠C = ∠F and ΔABC ∼ ΔDEF

Given: In ΔABC and ΔDEF,    AB DE =  BC EF =  AC DF .

To Prove: ∠A = ∠D, ∠B = ∠E and ∠C = ∠F

Construction: Draw PQ, Such that DP = AB and DQ = AC.

Proof: In ΔABC and ΔDEF,

AB DE =  AC DF (Given)

DP DE =  DQ DF (By Construction)

DE DP =  DF DQ (Taking Reciprocals)

DE DP - 1 =  DF DQ - 1 (Substracting 1 from both sides)

DE-DP DP =  DF-DQ DQ

PE DP =  QF DQ

DP PE =  DQ QF (Taking Reciprocals)

PQ ∥ EF (By converse of basic proportionality theorem)

∠P = ∠E and ∠Q = ∠F (Pair of corresponding angles)

DP PE =  DQ QF =  PQ EF

AB DE =  AC DF =  BC EF (Given)

∴   PQ EF =  BC EF

⇒ BC = PQ

SAS similarity theorem or criteria:
If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the triangles are similar.

In ΔABC and ΔDEF,  AB DE =  AC DF and ∠A = ∠D then  ΔABC ∼ ΔDEF.

Given: In ΔABC and ΔDEF, AB/DE = AC/DF and ∠A = ∠D.

To prove: ΔABC ~  ΔDEF

Construction: Draw PQ such that DP = AB and DQ = AC.

Proof:

AB DE =  AC DF

DP DE =  DQ DF

DE DP =  DF DQ ( Taking reciprocal)

DP + PE DP =  DQ + QF DQ

DP DP +  PE DP = DQ DQ +  QF DQ

1 +  PE DP = 1 +  QF DQ

PE DP =  QF DQ

DP PE =  DQ QF

PQ ∥ EF (By converse of basic proportionally theorem)

∠P = ∠E, ∠Q = ∠F

In ΔDPQ and ΔDEF,

Hence ∠P = ∠E and ∠Q = ∠F

∴ ΔDPQ ~ ΔDEF (AA similarity criterion) ..... (1)

In ΔABC and ΔDPQ,

AB = DP, AC = DQ, ∠A = ∠D

ΔABC ≅ ΔDPQ (By SAS axiom)

ΔABC ~ ΔDPQ ......... (2)

From (1) and (2),

ΔABC ~  ΔDEF

AA similarity criterion:
When two corresponding angles of two triangles are equal, then the two triangles are similar. This is referred to as the AA similarity criterion.