Prove that if in two triangles,two angles and the included side of one triangle are equal to two angles and the included side of the other triangle,then the two triangles are congruent.
(N/A) This is the $ASA$ (Angle-Side-Angle) congruence criterion.
Let there be two triangles $\triangle ABC$ and $\triangle DEF$ such that $\angle B = \angle E$,$\angle C = \angle F$,and $BC = EF$.
We need to prove that $\triangle ABC \cong \triangle DEF$.
Case $1$: If $AB = DE$,then $\triangle ABC \cong \triangle DEF$ by $SAS$ congruence rule.
Case $2$: If $AB < DE$,take a point $P$ on $DE$ such that $DP = AB$. Join $PF$. In $\triangle ABC$ and $\triangle DPF$,$AB = DP$,$\angle B = \angle E$,and $BC = EF$. Thus,$\triangle ABC \cong \triangle DPF$ by $SAS$. This implies $\angle ACB = \angle DPF$. But we are given $\angle ACB = \angle DFE$. Therefore,$\angle DPF = \angle DFE$,which is only possible if $P$ coincides with $E$. Thus,$AB = DE$ and $\triangle ABC \cong \triangle DEF$.
Case $3$: If $AB > DE$,a similar argument shows $AB = DE$ and the triangles are congruent.
Let there be two triangles $\triangle ABC$ and $\triangle DEF$ such that $\angle B = \angle E$,$\angle C = \angle F$,and $BC = EF$.
We need to prove that $\triangle ABC \cong \triangle DEF$.
Case $1$: If $AB = DE$,then $\triangle ABC \cong \triangle DEF$ by $SAS$ congruence rule.
Case $2$: If $AB < DE$,take a point $P$ on $DE$ such that $DP = AB$. Join $PF$. In $\triangle ABC$ and $\triangle DPF$,$AB = DP$,$\angle B = \angle E$,and $BC = EF$. Thus,$\triangle ABC \cong \triangle DPF$ by $SAS$. This implies $\angle ACB = \angle DPF$. But we are given $\angle ACB = \angle DFE$. Therefore,$\angle DPF = \angle DFE$,which is only possible if $P$ coincides with $E$. Thus,$AB = DE$ and $\triangle ABC \cong \triangle DEF$.
Case $3$: If $AB > DE$,a similar argument shows $AB = DE$ and the triangles are congruent.
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