If two angles and the included side of one triangle are equal to | Math Question and Answer

Answer:

Step by Step Explanation:

Let $△ A B C$ and $△ D E F$ be the two triangles such that $B C = E F, ∠ A C B = ∠ D F E,$ and $∠ A B C = ∠ D E F$ .
We need to prove that $\triangle ABC \cong \triangle DEF$ .
Let us assume that $A C = D F$ .

In $△ A B C$ and $△ D E F$ , we have $\begin{aligned} A C = D F & [Just assumed] \\ B C = E F & [By step 1] \\ ∠ A C B = ∠ D F E & [By step 1] \\ ∴ △ A B C ≅ △ D E F & [By SAS-Criterion] \end{aligned}$ Now, let us assume $A C \neq D F$ .

Let us construct D' on the line $F D$ such that $D^{'} F = A C$ and then join the point $D^{'}$ to the point $E$ .
In $△ A B C$ and $△ D^{'} E F$ , we have $\begin{aligned} A C = D^{'} F & [By construction] \\ B C = E F & [By step 1] \\ ∠ A C B = ∠ D F E & [By step 1] \\ ∴ △ A B C ≅ △ D^{'} E F & [By SAS-criterion] \end{aligned}$
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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