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Corollary (AAS Congruence Criterion)

Theorem: If any two angles and a non-included side of one triangle are equal to the corresponding angles and side of another triangle, then the two triangles are congruent.

Given: In ΔABC and ΔDEF
A = D
B = E
BC = EF


To Prove: ΔABC ΔDEF

Proof:
Since the sum of the angles of a triangle is 180°, we have
A + B + C = D + E + F
Since A = D and B = E, ... (Given)
C = F ... (i)
Now, In ΔABC and ΔDEF,
B = E ... (Given)
C = F ... [From (i)]
BC = EF ... (Given)
ΔABC
ΔDEF ... (ASA Cong. theorem.)
 

Example 1 : PR and QT are perpendicular to PQ and PR = QT. Show that AP = AQ.
 


In ∆PAR and ∆QAT

RPA = TQA (Each is 90°)

PAR = QAT (Vertically opposite angles)

and PR = QT

∆PAR

∆QAT (AAS)

AP = AQ.

 

 

Example 2 : In the figure, it is given that RT = TS, 1 = 2 2 and 4 = 2 3 prove that Δ RBT Δ SAT

In Δ RTS, we have

ST = RT

TRS = TSR.. ….(i)

We have,

1 = 4[vertically opposite angles]

2 2 = 2 3[ 1 = 2 2 and 4 = 2 3 (given)]

2 = 3………(ii)

Subtracting (ii) from (i), we get (ii)

TRS - 2 = TSR - 3

TRB = TSA(iii)

Thus, in triangles RBT and SAT, we have

RTB = STA[common]

RT = ST [given]

And, TRB = TSA[from (iii)]

So, by ASA congruence criterion, we have

Δ RBT SAT





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