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, then prove that the two triangles are congruent.


Answer:


Step by Step Explanation:
  1. Let ABC and DEF be the two triangles such that BC=EF,ACB=DFE, and ABC=DEF.
      A C B D F E
  2. We need to prove that ABCDEF.
  3. Let us assume that AC=DF.

    In ABC and DEF, we have AC=DF  [Just assumed]BC=EF  [By step 1]ACB=DFE  [By step 1] Now, let us assume AC \neq DF .

    Let us construct D' on the line FD such that D'F = AC and then join the point D' to the point E .
      A C B D F ED' D'
    In \triangle ABC and \triangle D'EF , we have \begin{aligned} AC = D'F & {\space} {\space} \text{[By construction]} \\ BC = EF & {\space} {\space} \text{[By step 1]} \\ \angle ACB = \angle DFE & {\space} {\space} \text{[By step 1]} \\ \therefore \triangle ABC \cong \triangle D'EF & {\space} {\space} \text{[By SAS-criterion]} \\ \end{aligned}
  4. As corresponding parts of congruent triangles are equal, we have \angle ABC = \angle D'EF But, \angle ABC = \angle DEF \space \text{[Given]}

    \implies \angle D'EF = \angle DEF

    This is possible only when D and D' coincide.

    Hence, \bf {\triangle ABC \cong \triangle DEF}.

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