# 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 Î”ABC and Î”DEF

âˆ A = âˆ D

âˆ B = âˆ E

BC = EF

**To Prove:Â **Î”ABC â‰… Î”DEF

**Proof:**

Since the sum of the angles of a triangle is 180Â°, we have

âˆ A + âˆ B + âˆ C = âˆ D + âˆ E + âˆ F

Since âˆ A = âˆ D and âˆ B = âˆ E, ... (Given)

âˆ C = âˆ F ... (i)

Now, In Î”ABC and Î”DEF,

âˆ B = âˆ E ... (Given)

âˆ C = âˆ F ... [From (i)]

BC = EF ... (Given)

âˆ´ Î”ABC â‰… Î”DEF ... (ASA Cong. theorem.)

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

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In âˆ†PAR and âˆ†QAT

âˆ RPA = âˆ TQA (Each is 90Â°)

âˆ PAR = âˆ QAT (Vertically opposite angles)

and PR = QT

âˆ´ âˆ†PAR

â‰… âˆ†QAT (AAS)âˆ´ AP = AQ.

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**Example 2 :** In the figure, it is given that RT = TS, âˆ 1 = 2âˆ 2 and âˆ 4 = 2âˆ 3 prove that Î” RBT â‰… Î” SAT

In Î” RTS, we have

ST = RT

â‡’ âˆ TRS =âˆ TSR.. â€¦.(i)

We have,

âˆ 1 = âˆ 4[vertically opposite angles]

â‡’ 2 âˆ 2 = 2âˆ 3[âˆ 1 = 2âˆ 2 and âˆ 4 = 2âˆ 3 (given)]

â‡’ âˆ 2 = âˆ 3â€¦â€¦â€¦(ii)

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

âˆ TRS - âˆ 2 = âˆ TSR - âˆ 3

â‡’ âˆ TRB = âˆ TSA(iii)

Thus, in triangles RBT and SAT, we have

âˆ RTB = âˆ STA[common]

RT = ST [given]

And, âˆ TRB = âˆ TSA[from (iii)]

So, by ASA congruence criterion, we have

Î” RBT â‰… âˆ SAT