In the previous section we have learned that congruent triangles completely and exactly cover each other. We have also learned that the correspondence is also important for the congruence to work. In this section we will discuss about the 'Criteria for the Congruence of Triangles'.
Fig.10.8(a) below, shows a triangle ABC. Fig.(b) shows five triangles: XYZ, PQR, UVW, MNP and NOP.
Our problem is this: Is there any triangle in fig.(b), which is congruent to ABC? If yes, which one?
Solution: Looking at fig.(a) & (b), we find that the measurements of all the 3 sides of all the triangles are given. So here we use a rule known as the SSS criterion. The ‘S’ stands for ‘Side’. So this is the abbreviation for 'Side Side Side criterion'. The 3 sides of a triangle.
■ This rule is based on the fact that, when two triangles are congruent, their 3 sides will be equal. Let us apply this rule to our problem:
• In fig.(a), We have ΔABC with the 3 sides: AB = 3.0 cm, BC = 2.6 cm and CA = 4.1 cm
• In fig.(b) we have one particular triangle, which is ΔUVW with the 3 sides: UV = 3.0 cm, VW = 4.1 cm and WU = 2.6 cm
• The two triangles are having the same sides. So they are congruent.
But the order of the sides is not correct. It is our job to specify the correct order. We do this by writing the correspondence. How do we write the correspondence with out making a ‘trace copy of UVW’, and placing it over ABC? It can be done as follows:
• Consider ABC. Take any two sides say 3.0 cm and 2.6 cm. They meet at B
• Corresponding to this, in UVW, 3.0 cm and 2.6 cm meet at U
• So we see that B corresponds to U That is., B↔U
• Now, starting from B, segment BA is 3.0. And starting from U, segment UV is 3.0
• So we see that A corresponds to V That is., A↔V
• The only ones remaining are C and W, and obviously, they will correspond to each other because they are the only remaining ones. So we get C↔W
• So we write: A↔V, B↔U and C↔W
• This is same as writing: ABC↔ VUW.
Note that, the name of the second triangle is UVW, but in correspondence, it is VUW. We can write: ΔABC ≅ ΔUVW under the correspondence: ABC↔VUW.
The animation given below in fig.10.9, shows the superposition and the correspondence:
Based on the above discussion, we can write down the criterion:
■ SSS Congruence criterion:
If under a given correspondence, the three sides of one triangle are equal to the three corresponding sides of another triangle, then the triangles are congruent.
Solved example 10.4
In the fig.10.10 given below, prove that, ΔPRS and ΔPRQ are congruent. Also state the relation between the ∠SPR and ∠QPR
Solution:
Part 1:
• Take ΔPRS. Consider the side PS. It is shown to be equal to the side PQ of the other ΔPRQ
• Consider the side SR. It is shown to be equal to RQ of the other ΔPRQ
• PR is common to both the triangles. So the third side PR of ΔPRS is equal to the third side PR of ΔPRQ
• So all the 3 sides are same. The triangles are congruent
• But we have to write the correspondence:
• The equal sides PS and PQ start from P. So P↔P and S↔Q.
• The only remaining vertex in both the triangles is R. So R↔R
• So we can write: P↔P, S↔Q, and R↔R. This is same as PSR↔PQR
• Thus we can write: ΔPRS ≅ ΔPQR under the correspondence: PSR↔PQR
Part 2:
• From part 1, we have seen that the two triangles are congruent. That is., they are exact copies. So the angles at the corresponding corners must also be the same.
• ∠SPR is the angle at P in the ΔPRS
• ∠QPR is the angle at P in the ΔPQR
• We have seen that P↔P. So the angle at P will be same in both the triangles
• Thus the relation is that: ∠SPR and ∠QPR are equal
Solved example 10.5
In the fig.10.11 below, if PM = ON and PN = OM, which of the following is true?
(i) ΔPMN≅ΔMNO (ii) ΔPNM≅ΔOMN (iii) ΔPMN≅ΔOMN (iv) ΔPMN≅ΔONM
Solution:
• We have two triangles: ΔMNP and ΔMNO
• Given PM = ON and PN = OM. The third side MN is common to both the triangles. So all 3 sides are the same.
• Thus ΔMNP and ΔMNO are congruent to one another. But we have to write the correspondence.
• In ΔMNP, take the two sides: PM and PN. They meet at P
• In ΔMNO, the corresponding sides are ON and OM respectively. These two sides meet at O. So we get P↔O
• Consider PM in ΔPMN. It starts from P and moves towards M. The corresponding side ON in ΔMNO starts from O and moves towards N. Thus we get M↔N
• All statements which satisfy P↔O and M↔N are true
(i) ΔPMN≅ΔMNO is not true because, in this statement, P↔M, which is not true
(ii) ΔPNM≅ΔOMN is true because, in this statement, P↔O, N↔M, M↔N, are all true
(iii) ΔPMN≅ΔOMN is not true because, M↔M and N↔N is not true
(iv) ΔPMN≅ΔONM is true because, P↔O, M↔N, N↔M, are all true
So we have completed the discussion on SSS criterion. In the next section we will discuss about another criterion for the Congruence of Triangles.
Fig.10.8(a) below, shows a triangle ABC. Fig.(b) shows five triangles: XYZ, PQR, UVW, MNP and NOP.
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| Fig.10.8 |
Solution: Looking at fig.(a) & (b), we find that the measurements of all the 3 sides of all the triangles are given. So here we use a rule known as the SSS criterion. The ‘S’ stands for ‘Side’. So this is the abbreviation for 'Side Side Side criterion'. The 3 sides of a triangle.
■ This rule is based on the fact that, when two triangles are congruent, their 3 sides will be equal. Let us apply this rule to our problem:
• In fig.(a), We have ΔABC with the 3 sides: AB = 3.0 cm, BC = 2.6 cm and CA = 4.1 cm
• In fig.(b) we have one particular triangle, which is ΔUVW with the 3 sides: UV = 3.0 cm, VW = 4.1 cm and WU = 2.6 cm
• The two triangles are having the same sides. So they are congruent.
But the order of the sides is not correct. It is our job to specify the correct order. We do this by writing the correspondence. How do we write the correspondence with out making a ‘trace copy of UVW’, and placing it over ABC? It can be done as follows:
• Consider ABC. Take any two sides say 3.0 cm and 2.6 cm. They meet at B
• Corresponding to this, in UVW, 3.0 cm and 2.6 cm meet at U
• So we see that B corresponds to U That is., B↔U
• Now, starting from B, segment BA is 3.0. And starting from U, segment UV is 3.0
• So we see that A corresponds to V That is., A↔V
• The only ones remaining are C and W, and obviously, they will correspond to each other because they are the only remaining ones. So we get C↔W
• So we write: A↔V, B↔U and C↔W
• This is same as writing: ABC↔ VUW.
Note that, the name of the second triangle is UVW, but in correspondence, it is VUW. We can write: ΔABC ≅ ΔUVW under the correspondence: ABC↔VUW.
The animation given below in fig.10.9, shows the superposition and the correspondence:
 |
| Fig.10.9 |
■ SSS Congruence criterion:
If under a given correspondence, the three sides of one triangle are equal to the three corresponding sides of another triangle, then the triangles are congruent.
Solved example 10.4
In the fig.10.10 given below, prove that, ΔPRS and ΔPRQ are congruent. Also state the relation between the ∠SPR and ∠QPR
 |
| Fig.10.10 |
Part 1:
• Take ΔPRS. Consider the side PS. It is shown to be equal to the side PQ of the other ΔPRQ
• Consider the side SR. It is shown to be equal to RQ of the other ΔPRQ
• PR is common to both the triangles. So the third side PR of ΔPRS is equal to the third side PR of ΔPRQ
• So all the 3 sides are same. The triangles are congruent
• But we have to write the correspondence:
• The equal sides PS and PQ start from P. So P↔P and S↔Q.
• The only remaining vertex in both the triangles is R. So R↔R
• So we can write: P↔P, S↔Q, and R↔R. This is same as PSR↔PQR
• Thus we can write: ΔPRS ≅ ΔPQR under the correspondence: PSR↔PQR
Part 2:
• From part 1, we have seen that the two triangles are congruent. That is., they are exact copies. So the angles at the corresponding corners must also be the same.
• ∠SPR is the angle at P in the ΔPRS
• ∠QPR is the angle at P in the ΔPQR
• We have seen that P↔P. So the angle at P will be same in both the triangles
• Thus the relation is that: ∠SPR and ∠QPR are equal
Solved example 10.5
In the fig.10.11 below, if PM = ON and PN = OM, which of the following is true?
(i) ΔPMN≅ΔMNO (ii) ΔPNM≅ΔOMN (iii) ΔPMN≅ΔOMN (iv) ΔPMN≅ΔONM
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| Fig.10.11 |
• We have two triangles: ΔMNP and ΔMNO
• Given PM = ON and PN = OM. The third side MN is common to both the triangles. So all 3 sides are the same.
• Thus ΔMNP and ΔMNO are congruent to one another. But we have to write the correspondence.
• In ΔMNP, take the two sides: PM and PN. They meet at P
• In ΔMNO, the corresponding sides are ON and OM respectively. These two sides meet at O. So we get P↔O
• Consider PM in ΔPMN. It starts from P and moves towards M. The corresponding side ON in ΔMNO starts from O and moves towards N. Thus we get M↔N
• All statements which satisfy P↔O and M↔N are true
(i) ΔPMN≅ΔMNO is not true because, in this statement, P↔M, which is not true
(ii) ΔPNM≅ΔOMN is true because, in this statement, P↔O, N↔M, M↔N, are all true
(iii) ΔPMN≅ΔOMN is not true because, M↔M and N↔N is not true
(iv) ΔPMN≅ΔONM is true because, P↔O, M↔N, N↔M, are all true
So we have completed the discussion on SSS criterion. In the next section we will discuss about another criterion for the Congruence of Triangles.
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