In the previous section we have learned the SAS Criterion for the congruence of Triangles. In this section we will discuss about another criterion.
Fig.10.12(a) below, shows a triangle ABC. Fig.(b) shows five triangles: XYZ, WYZ, PQR, DEF, and MNO.
Our problem is this: Is there any triangle in fig.(b), which is congruent to ΔABC? If yes, which one?
Solution: Looking at figs.(a) & (b), we find that the measurements of the triangles are incomplete. None of the triangles have all lengths of all sides, or angles at all corners, marked on them.
• If the triangles had the lengths of all the 3 sides, then we could straight away use the SSS criterion to check the congruence.
• If two sides and their included angle is given, we could use the SAS criterion.
But in this situation, these are not possible because the measurements are incomplete. We cannot find an SSS comparison or SAS comparison. In such situations, we must try to use other criteria.
We are going to use a rule known as the ASA criterion. As before, the 'A' stands for angle, and, the ‘S’ stands for ‘Side’. So this is the abbreviation for 'Angle Side Angle criterion'. Two angles and one side.
■ This rule is based on the fact that, when two triangles are congruent, any two angles, and the side between those angles, in one triangle will be present in the other triangle. This rule will become clear when we apply it to our problem:
• In fig.(a), We have ΔABC with two angles and the included side: ∠A = 65o, ∠C = 82o and the included side AC = 2.7 cm
• In fig.(b) we have one particular triangle, which is ΔDEF with the two angles and the included side: ∠E = 65o, ∠D = 82o and the included side ED = 2.7 cm
• The two angles and the included side are the same. So ΔABC and ΔDEF are congruent.
■ We are able to establish congruence just by using two angles and the included side. All other sides and angles may or may not be given in the problem. But we do not need them to establish congruence.
Some important points to note:
• 'ASA' denotes two angles and one side. This side should be the included side between the chosen angles. If we take any other side, the congruence will not work.
• In this problem, we chose the angles ∠A and ∠C, and the side AC between them. We used them for the comparison, and arrived at the conclusion that ΔABC and ΔDEF are in congruence.
• If ΔABC and ΔDEF are in congruence, there will surely be other two ASA combinations also. We will write all the three:
■ [∠A, ∠C, and the included side AC] has a corresponding combination which we already found out: [∠E, ∠D, and the included side ED]
■ [∠B, ∠C, and the included side BC] will have a corresponding combination, which we are yet to find
■ [∠A, ∠B, and the included side AB] will have a corresponding combination, which we are yet to find
So there are two details that we are yet to find. Those two details are not necessary to establish a congruence. But establishing a congruence does not solve the problem completely. We have a little more work to do. We have to put the sides and corners of the two triangles in order. We do this by writing the correspondence. And, after writing the correspondence, we will be able to write those two details very easily.
So let us try to write the correspondence:
The one detail which we already know, can be shown by a rough sketch as in the fig.10.22 below:
• From the fig.10.22, it is obvious that A↔E and C↔D. Because, ∠A and ∠E have the same measure, and similarly, ∠C and ∠D have the same measure
• The only remaining corner in the first triangle is B, and that in the second triangle is F. So we get B↔F
• So the correspondence is: A↔E, B↔F, and C↔D. This is same as ABC↔EFD
• Thus we can write: ΔABC and ΔDEF are congruent to one another under the correspondence ABC↔EFD
• This is same as writing: ΔABC ≅ ΔEFD
Now, we use the rough sketch in fig.10.13 above to write the two missing details:
■ [∠B, ∠C, and the included side BC] in ΔABC has a corresponding combination in ΔDEF. What is it?
• We have B↔F, and C↔D. So BC↔FD
• Thus the corresponding combination is: [∠F, ∠D, and the included side FD]
■ [∠A, ∠B, and the included side AB] in ΔABC has a corresponding combination in ΔDEF. What is it?
• We have A↔E, and B↔F. So AB↔EF
• Thus the corresponding combination is: [∠E, ∠F, and the included side EF]
So we have established the congruence, and put every relations between the two triangles in order. The following animation shows the superposition of ΔDEF over ΔABC:
Based on the above discussion, we can write down the criterion:
■ SAS Congruence criterion:
If under a correspondence, 'two angles and the included side' of a triangle are equal to 'two corresponding angles and the included side' of another triangle, then the triangles are congruent.
Solved example 10.9
There are two triangles. ΔPQR and ΔXYZ. We are required to establish a congruence: ΔPQR ≅ ΔYXZ. It is known that PQ = YX. What additional information is required to establish the congruence?
Solution:
• The required congruence is: ΔPQR ≅ ΔYXZ
• So P↔Y, Q↔X and R↔Z
• It is known that PQ = YX
• Let us draw a rough sketch as shown below:
• From the fig., it is clear that, if we are given an additional information that, ∠P = ∠Y, and ∠Q = ∠X, the two triangles will be congruent (ASA congruence)
Solved example 10.10
In fig.10.25, UW = VX. Establish the congruence between ΔOUW and ΔOVX, and write the correspondence
Solution:
• In the two triangles, UW = VX. So we have one side equal.
• If we can prove that the 'corresponding angles at the ends of these sides' also equal, then the triangles will be congruent by the ASA rule. So let us try to prove it:
• We have ∠U + ∠O = 27 + 37 = 64o
• ∴ ∠W = 180 - 64 = 116o (∵ sum of the interior angles of a triangle is 180o)
• ∠UOW = ∠VOX = 37o (∵ they are opposite angles )
• In ΔOVX, ∠O + ∠V = 37 + 27 = 64o
• ∴ ∠X = 180 - 64 = 116o (∵ sum of the interior angles of a triangle is 180o)
• Thus we get ∠W = ∠X. We already have ∠U = ∠V and lengths of sides UW = XV
• So ΔOUW and ΔOXV are congruent according to the ASA congruence criterion
• Now we have to write the correspondence:
• We have ∠W = ∠X and ∠U = ∠V
• So W↔X and U↔V. The only remaining corner is O. So O↔O
• This can be written as OUW↔OVX
• So we can write: ΔOUW and ΔOXV are congruent under the correspondence OUW↔OVX
• This can also be written as ΔOUW ≅ ΔOVX
Solved example 10.11
Given below are some pairs of triangles. In each pair, examine if the triangles are congruent to one another. If they are congruent, write the correspondence.
(i) ΔABC: ∠A = 30o, ∠B = 45o, AB = 4 cm. │ΔXYZ: ∠Y = 30o, ∠Z = 45o, YZ = 4 cm
(ii) ΔPQR: ∠P = 40o, ∠Q = 60o, PQ = 7 cm. │ΔABC: ∠A = 40o, ∠B = 60o, AC = 7 cm
(iii) ΔXYZ: ∠X = 80o, ∠Y = 50o, XY = 6 cm. │ΔPQR: ∠P = 80o, ∠R = 50o, PQ = 6 cm
Solution:
(i) A rough sketch is shown in the fig.10.26(i)
• Side AB = YZ, ∠A = ∠Y and ∠B = ∠Z. So the triangles are congruent by ASA criterion
• Now we write the correspondence:
• ∠A = ∠Y = 30o. So A↔Y.
• ∠B = ∠Z = 45o. So B↔Z
• The remaining corners are C and X. So C↔X.
• Thus we have A↔Y, B↔Z, C↔X ⇒ ABC↔YZX
• So ΔABC and ΔXYZ are congruent under the correspondence ABC↔YZX
• This is same as ΔABC ≅ ΔYZX
(ii) A rough sketch is shown in the fig.10.26(ii)
• ∠P = ∠A and ∠Q = ∠B. But length of AB is not given.
• AC is given as 7 cm. If ∠C = 60o, AC will correspond to PQ
• So we have to find ∠C:
• ∠C = 180 - (40 + 60) (∵ sum of the interior angles of a triangle is 180o)
• So ∠C = 180 - 100 = 80o
• So ∠C is not 60o, and thus, without the length of AB, we can not say if the two triangles are congruent or not
(iii) A rough sketch is shown in the fig.10.26(iii)
• ∠X = ∠P and ∠Y = ∠R. But length of PR is not given.
• PQ is given as 7 cm. If ∠Q = 50o, PQ will correspond to XY
• So we have to find ∠Q:
• ∠Q = 180 - (80 + 50) (∵ sum of the interior angles of a triangle is 180o)
• So ∠Q = 180 - 130 = 50o
Q is indeed equal to 50. So we have an ASA congruence
• Now we write the correspondence:
• ∠X = ∠P = 80o. So X↔P.
• ∠Z = ∠Q = 50o. So Z↔Q
• The remaining corners are Y and R. So Y↔R.
• Thus we have X↔P, Y↔R, Z↔Q ⇒ XYZ↔PRQ
• So ΔXYZ and ΔPQR are congruent under the correspondence XYZ↔PRQ
• This is same as ΔXYZ ≅ ΔPRQ
Solved example 10.12
In the fig.10.27 below, the ray OS bisects ∠POR. It also bisects ∠PQR. Prove that OP = OR and PQ = RQ
Solution:
• The ray OS bisects ∠ROP. So we get ∠ROQ = ∠POQ
• The ray OS bisects ∠PQR. So we get ∠PQO = ∠RQO
• OQ is common. So we have an ASA congruence
• The triangles ΔOQR and ΔOQP are congruent.
• Now we have to write the correspondence:
■ For the ASA congruence, we took [∠ROQ, ∠OQR, and their included side OQ] in ΔOQR
■ We took [∠POQ, ∠OQP and their included side OQ] in ΔOQP
• The above two are the corresponding combinations
• In those corresponding combinations, ∠ROQ = ∠POQ. So we get O↔O
• Also ∠OQR = ∠OQP. So we get Q↔Q
• The only remaining corners are P and R So we get P↔R
• Thus we can write OPQ↔ORQ
• So the two triangles ΔOPQ and ΔOQR are congruent under the correspondence: OPQ↔ORQ
• This is same as ΔOPQ ≅ ΔORQ
• Now we have to prove that OP = OR
• We have already proved that O↔O and P↔R
• So OP and OR are corresponding sides. Corresponding sides of congruent triangles will be equal in length. Thus OP = OR
• Next we have to prove that PQ = RQ
• We have already proved that P↔R and Q↔Q
So PQ and RQ are corresponding sides. Corresponding sides of congruent triangles will be equal in length. Thus PQ = RQ
So we have completed the discussion on ASA criterion. In the next section we will discuss about one more criterion.
Fig.10.12(a) below, shows a triangle ABC. Fig.(b) shows five triangles: XYZ, WYZ, PQR, DEF, and MNO.
 |
| Fig.10.21 |
Solution: Looking at figs.(a) & (b), we find that the measurements of the triangles are incomplete. None of the triangles have all lengths of all sides, or angles at all corners, marked on them.
• If the triangles had the lengths of all the 3 sides, then we could straight away use the SSS criterion to check the congruence.
• If two sides and their included angle is given, we could use the SAS criterion.
But in this situation, these are not possible because the measurements are incomplete. We cannot find an SSS comparison or SAS comparison. In such situations, we must try to use other criteria.
We are going to use a rule known as the ASA criterion. As before, the 'A' stands for angle, and, the ‘S’ stands for ‘Side’. So this is the abbreviation for 'Angle Side Angle criterion'. Two angles and one side.
■ This rule is based on the fact that, when two triangles are congruent, any two angles, and the side between those angles, in one triangle will be present in the other triangle. This rule will become clear when we apply it to our problem:
• In fig.(a), We have ΔABC with two angles and the included side: ∠A = 65o, ∠C = 82o and the included side AC = 2.7 cm
• In fig.(b) we have one particular triangle, which is ΔDEF with the two angles and the included side: ∠E = 65o, ∠D = 82o and the included side ED = 2.7 cm
• The two angles and the included side are the same. So ΔABC and ΔDEF are congruent.
■ We are able to establish congruence just by using two angles and the included side. All other sides and angles may or may not be given in the problem. But we do not need them to establish congruence.
Some important points to note:
• 'ASA' denotes two angles and one side. This side should be the included side between the chosen angles. If we take any other side, the congruence will not work.
• In this problem, we chose the angles ∠A and ∠C, and the side AC between them. We used them for the comparison, and arrived at the conclusion that ΔABC and ΔDEF are in congruence.
• If ΔABC and ΔDEF are in congruence, there will surely be other two ASA combinations also. We will write all the three:
■ [∠A, ∠C, and the included side AC] has a corresponding combination which we already found out: [∠E, ∠D, and the included side ED]
■ [∠B, ∠C, and the included side BC] will have a corresponding combination, which we are yet to find
■ [∠A, ∠B, and the included side AB] will have a corresponding combination, which we are yet to find
So there are two details that we are yet to find. Those two details are not necessary to establish a congruence. But establishing a congruence does not solve the problem completely. We have a little more work to do. We have to put the sides and corners of the two triangles in order. We do this by writing the correspondence. And, after writing the correspondence, we will be able to write those two details very easily.
So let us try to write the correspondence:
The one detail which we already know, can be shown by a rough sketch as in the fig.10.22 below:
 |
| Fig.10.22 |
• The only remaining corner in the first triangle is B, and that in the second triangle is F. So we get B↔F
• So the correspondence is: A↔E, B↔F, and C↔D. This is same as ABC↔EFD
• Thus we can write: ΔABC and ΔDEF are congruent to one another under the correspondence ABC↔EFD
• This is same as writing: ΔABC ≅ ΔEFD
Now, we use the rough sketch in fig.10.13 above to write the two missing details:
■ [∠B, ∠C, and the included side BC] in ΔABC has a corresponding combination in ΔDEF. What is it?
• We have B↔F, and C↔D. So BC↔FD
• Thus the corresponding combination is: [∠F, ∠D, and the included side FD]
■ [∠A, ∠B, and the included side AB] in ΔABC has a corresponding combination in ΔDEF. What is it?
• We have A↔E, and B↔F. So AB↔EF
• Thus the corresponding combination is: [∠E, ∠F, and the included side EF]
So we have established the congruence, and put every relations between the two triangles in order. The following animation shows the superposition of ΔDEF over ΔABC:
 |
| Fig.10.23 |
■ SAS Congruence criterion:
If under a correspondence, 'two angles and the included side' of a triangle are equal to 'two corresponding angles and the included side' of another triangle, then the triangles are congruent.
Solved example 10.9
There are two triangles. ΔPQR and ΔXYZ. We are required to establish a congruence: ΔPQR ≅ ΔYXZ. It is known that PQ = YX. What additional information is required to establish the congruence?
Solution:
• The required congruence is: ΔPQR ≅ ΔYXZ
• So P↔Y, Q↔X and R↔Z
• It is known that PQ = YX
• Let us draw a rough sketch as shown below:
 |
| Fig.10.24 |
Solved example 10.10
In fig.10.25, UW = VX. Establish the congruence between ΔOUW and ΔOVX, and write the correspondence
 |
| Fig.10.25 |
• In the two triangles, UW = VX. So we have one side equal.
• If we can prove that the 'corresponding angles at the ends of these sides' also equal, then the triangles will be congruent by the ASA rule. So let us try to prove it:
• We have ∠U + ∠O = 27 + 37 = 64o
• ∴ ∠W = 180 - 64 = 116o (∵ sum of the interior angles of a triangle is 180o)
• ∠UOW = ∠VOX = 37o (∵ they are opposite angles )
• In ΔOVX, ∠O + ∠V = 37 + 27 = 64o
• ∴ ∠X = 180 - 64 = 116o (∵ sum of the interior angles of a triangle is 180o)
• Thus we get ∠W = ∠X. We already have ∠U = ∠V and lengths of sides UW = XV
• So ΔOUW and ΔOXV are congruent according to the ASA congruence criterion
• Now we have to write the correspondence:
• We have ∠W = ∠X and ∠U = ∠V
• So W↔X and U↔V. The only remaining corner is O. So O↔O
• This can be written as OUW↔OVX
• So we can write: ΔOUW and ΔOXV are congruent under the correspondence OUW↔OVX
• This can also be written as ΔOUW ≅ ΔOVX
Solved example 10.11
Given below are some pairs of triangles. In each pair, examine if the triangles are congruent to one another. If they are congruent, write the correspondence.
(i) ΔABC: ∠A = 30o, ∠B = 45o, AB = 4 cm. │ΔXYZ: ∠Y = 30o, ∠Z = 45o, YZ = 4 cm
(ii) ΔPQR: ∠P = 40o, ∠Q = 60o, PQ = 7 cm. │ΔABC: ∠A = 40o, ∠B = 60o, AC = 7 cm
(iii) ΔXYZ: ∠X = 80o, ∠Y = 50o, XY = 6 cm. │ΔPQR: ∠P = 80o, ∠R = 50o, PQ = 6 cm
Solution:
(i) A rough sketch is shown in the fig.10.26(i)
 |
| Fig.10.26 |
• Now we write the correspondence:
• ∠A = ∠Y = 30o. So A↔Y.
• ∠B = ∠Z = 45o. So B↔Z
• The remaining corners are C and X. So C↔X.
• Thus we have A↔Y, B↔Z, C↔X ⇒ ABC↔YZX
• So ΔABC and ΔXYZ are congruent under the correspondence ABC↔YZX
• This is same as ΔABC ≅ ΔYZX
(ii) A rough sketch is shown in the fig.10.26(ii)
• ∠P = ∠A and ∠Q = ∠B. But length of AB is not given.
• AC is given as 7 cm. If ∠C = 60o, AC will correspond to PQ
• So we have to find ∠C:
• ∠C = 180 - (40 + 60) (∵ sum of the interior angles of a triangle is 180o)
• So ∠C = 180 - 100 = 80o
• So ∠C is not 60o, and thus, without the length of AB, we can not say if the two triangles are congruent or not
(iii) A rough sketch is shown in the fig.10.26(iii)
• ∠X = ∠P and ∠Y = ∠R. But length of PR is not given.
• PQ is given as 7 cm. If ∠Q = 50o, PQ will correspond to XY
• So we have to find ∠Q:
• ∠Q = 180 - (80 + 50) (∵ sum of the interior angles of a triangle is 180o)
• So ∠Q = 180 - 130 = 50o
Q is indeed equal to 50. So we have an ASA congruence
• Now we write the correspondence:
• ∠X = ∠P = 80o. So X↔P.
• ∠Z = ∠Q = 50o. So Z↔Q
• The remaining corners are Y and R. So Y↔R.
• Thus we have X↔P, Y↔R, Z↔Q ⇒ XYZ↔PRQ
• So ΔXYZ and ΔPQR are congruent under the correspondence XYZ↔PRQ
• This is same as ΔXYZ ≅ ΔPRQ
Solved example 10.12
In the fig.10.27 below, the ray OS bisects ∠POR. It also bisects ∠PQR. Prove that OP = OR and PQ = RQ
 |
| Fig.10.27 |
• The ray OS bisects ∠ROP. So we get ∠ROQ = ∠POQ
• The ray OS bisects ∠PQR. So we get ∠PQO = ∠RQO
• OQ is common. So we have an ASA congruence
• The triangles ΔOQR and ΔOQP are congruent.
• Now we have to write the correspondence:
■ For the ASA congruence, we took [∠ROQ, ∠OQR, and their included side OQ] in ΔOQR
■ We took [∠POQ, ∠OQP and their included side OQ] in ΔOQP
• The above two are the corresponding combinations
• In those corresponding combinations, ∠ROQ = ∠POQ. So we get O↔O
• Also ∠OQR = ∠OQP. So we get Q↔Q
• The only remaining corners are P and R So we get P↔R
• Thus we can write OPQ↔ORQ
• So the two triangles ΔOPQ and ΔOQR are congruent under the correspondence: OPQ↔ORQ
• This is same as ΔOPQ ≅ ΔORQ
• Now we have to prove that OP = OR
• We have already proved that O↔O and P↔R
• So OP and OR are corresponding sides. Corresponding sides of congruent triangles will be equal in length. Thus OP = OR
• Next we have to prove that PQ = RQ
• We have already proved that P↔R and Q↔Q
So PQ and RQ are corresponding sides. Corresponding sides of congruent triangles will be equal in length. Thus PQ = RQ
So we have completed the discussion on ASA criterion. In the next section we will discuss about one more criterion.
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