Question

Prove the theorem if any two angles and non included side of One triangle are equal to the corresponding angle and side of another triangle then the two angles are congruent

Answer 1

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 < ED

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,