### 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.

Step by Step Explanation:
1. Let $\triangle ABC$ and $\triangle DEF$ be the two triangles such that $BC = EF, \angle ACB = \angle DFE,$ and $\angle ABC = \angle DEF$.
2. We need to prove that $\triangle ABC \cong \triangle DEF$.
3. Let us assume that $AC = DF$.

In $\triangle ABC$ and $\triangle DEF$, we have \begin{aligned} AC = DF & {\space} {\space} \text{[Just assumed]} \\ BC = EF & {\space} {\space} \text{[By step 1]} \\ \angle ACB = \angle DFE & {\space} {\space} \text{[By step 1]} \\ \therefore \triangle ABC \cong \triangle DEF & {\space} {\space} \text{[By SAS-Criterion]} \end{aligned} 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 $E$.
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, $\bf {\triangle ABC \cong \triangle DEF}.$ 