Math, asked by swapnama7, 1 year ago

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

pls send the answer fast...........................I will mark it as brainliest

Answers

Answered by Shaizakincsem
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 vikram991
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,
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