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We have dealt with congruence criteria in the previous classes. In this section, let us look at the axioms and theorems supporting various congruence criteria.

**The four main congruence criterion are**:

**1.**Side - Side - Side (SSS) congruence criterion

**2.**Side - Angle - Side (SAS) congruence criterion

**3.**Angle - Side - Angle (ASA) congruence criterion

**4.**Right angle - Hypotenuse - Side (RHS) congruence criterion

Axiom \(1\) (SAS congruence rule)

Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle.

Let us look at the figure given below.

From the figure given above, we can come up with the following inferences.

\(AB = PQ = 3 cm\)

\(BC = QR = 4 cm\)

\(\angle ABC = \angle PQR = 50^\circ\)

Here, two corresponding pairs of sides and one corresponding pairs of angles are equal.

Thus, \(\triangle ABC \cong \triangle PQR\) by the SAS congruence criterion.

Theorem \(1\) (ASA congruence rule)

Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle.

From the given statement, any two pair of corresponding angles and a pair of corresponding sides of the two triangles are required to satisfy the ASA congruence criterion.

**From the figure, let us try to prove that the two triangles are congruent to each other**.

To prove congruence in the two triangles, we should consider the following three cases.

Case \(1\): Assuming that \(AB = PQ\)

Now, the figure looks like this.

\(AB = PQ\)

**[Assumption]**\(\angle ABC = \angle PQR\)

**[Given]**\(BC = QR\)

**[Given]**Thus, by SAS congruence criterion, \(\triangle ABC \cong \triangle PQR\).

Case \(2\): Assuming \(AB > PQ\)

Now, mark a point \(E\) on \(AB\) such that \(EB\) equals \(PQ\).

The figure looks like this.

Now, \(EB = PQ\)

**[By construction]**\(\angle ABC = \angle PQR\)

**[Given]**\(BC = QR\)

**[Given]**Thus, by SAS congruence criterion, \(\triangle BEC \cong \triangle PQR\).

We have proved that the triangles are congruent to each other.

Hence, the other corresponding parts of the triangle will also be equal.

That is, \(\angle BEC = \angle QPR\), \(\angle ECB = \angle PRQ\), and \(EC = PR\).

Case \(3\): Assuming \(AB < PQ\)

Let us now mark a point \(M\) on \(PQ\) such that \(AB\) \(=\) \(MQ\) as shown in the figure given below.

Now, \(AB = QM\)

**[By construction]**\(\angle ABC = \angle MQR\)

**[Given]**\(BC = QR\)

**[Given]**Thus, by SAS congruence criterion, \(\triangle ABC \cong \triangle MQR\).

We have proved that the triangles are congruent to each other.

Hence, the other corresponding parts of the triangle will also be equal.

That is, \(\angle BAC = \angle QMR\), \(\angle ACB = \angle MRQ\), and \(EC = PR\).