prove that 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
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Answered by
211
Given: Two triangles ABC and DEF such that B = E, C = F
and BC = EF. To prove: ABC DEF
Proof: Case I: If AB = DE then in
ABC and DEF, AB = DE [by supposition] BC = EF [given] and B = E [given]
Thus, ABC DEF [SAS criterion]
Case II: If AB < DE Take a point G on ED such that EG = AB.
Join GF. In ABC and GEF, we have AB = GE [by supposition] BC = EF [given] B = E [given] Thus, ABC GEF [SAS criterion]
ACB = GFE [corresponding parts of congruent triangles are equal]
But ACB = DFE [given] GFE =DFE,
This is only possible when FG coincides with FD or G coincides with D.
AB must be equal to DE and hence, ABC DEF (by SAS) Case III: If AB > ED
With a similar argument (as in case II), we may conclude that ABC DEF (by SAS)
Thus, ABC DEF.
Answered by
187
here is your answer OK
Two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle.
Given: Two ΔsABC and DEF such that
B = E, C = F and BC = EF
To Prove:
Proof: There are three possibilities.
CASE I: When AB = DE,
In this case, we have
AB = DE
B = E [Given]
and, BC = EF [Given]
So, by SAS criterion of congruence, .
CASE II: When AB
In this case take a point G on ED such that EG = AB. Join GF.
Now, in ΔsABC and GEF, we have
AB = GE [By supposition]
B = E [Given]
and, BC = EF [Given]
So, by SAS criterion of congruence
⇒ ACB = GFE [ Corresponding parts of congruent triangles are equal]
But ACB = DFE [Given]
∴ GFE = DFE
This is possible only when ray FG coincides with ray FD or G coincides with D.
Thus, in ΔsABC and DEF, we have
AB = DE [As proved above]
B = E [Given]
and, BC = EF [Given]
So, by SAS criterion of congruent,
CASE III: When AB ED.
In this case take a point G on ED produced such that EG = AB. Join GF. Now, proceeding exactly on the same lines as in case II, we can prove that
Hence,
Two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle.
Given: Two ΔsABC and DEF such that
B = E, C = F and BC = EF
To Prove:
Proof: There are three possibilities.
CASE I: When AB = DE,
In this case, we have
AB = DE
B = E [Given]
and, BC = EF [Given]
So, by SAS criterion of congruence, .
CASE II: When AB
In this case take a point G on ED such that EG = AB. Join GF.
Now, in ΔsABC and GEF, we have
AB = GE [By supposition]
B = E [Given]
and, BC = EF [Given]
So, by SAS criterion of congruence
⇒ ACB = GFE [ Corresponding parts of congruent triangles are equal]
But ACB = DFE [Given]
∴ GFE = DFE
This is possible only when ray FG coincides with ray FD or G coincides with D.
Thus, in ΔsABC and DEF, we have
AB = DE [As proved above]
B = E [Given]
and, BC = EF [Given]
So, by SAS criterion of congruent,
CASE III: When AB ED.
In this case take a point G on ED produced such that EG = AB. Join GF. Now, proceeding exactly on the same lines as in case II, we can prove that
Hence,
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