In Delta ABC and Delta PQR,if frac{AB}{PQ} = | vedclass

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(A) Given that $\frac{AB}{PQ} = \frac{BC}{RQ} = \frac{AC}{PR} = 1$.This implies $AB = PQ$,$BC = RQ$,and $AC = PR$.By the $SSS$ (Side-Side-Side) congru

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(A) Given that $\frac{AB}{PQ} = \frac{BC}{RQ} = \frac{AC}{PR} = 1$.This implies $AB = PQ$,$BC = RQ$,and $AC = PR$.By the $SSS$ (Side-Side-Side) congruence criterion,if three sides of one triangle are equal to the three corresponding sides of another triangle,then the triangles are congruent.Matching the vertices based on the equality of sides:$AB = PQ$ (Vertex $A$ corresponds to $P$,$B$ to $Q$)$BC = RQ$ (Vertex $B$ corresponds to $R$,$C$ to $Q$)$AC = PR$ (Vertex $A$ corresponds to $P$,$C$ to $R$)Thus,$\Delta ABC \cong \Delta PQR$ is incorrect based on the ratio provided. Let's re-evaluate the correspondence: $A \leftrightarrow P$,$B \leftrightarrow Q$,$C \leftrightarrow R$ is not possible since $BC=RQ$ and $AC=PR$. Actually,$AB=PQ$,$BC=RQ$,$AC=PR$ implies $\Delta ABC \cong \Delta PQR$ is not the correct order. Let's check: $A$ corresponds to $P$,$B$ corresponds to $Q$,$C$ corresponds to $R$. Wait,$BC=RQ$ means $B$ corresponds to $R$ and $C$ corresponds to $Q$. $AC=PR$ means $A$ corresponds to $P$ and $C$ corresponds to $R$. $AB=PQ$ means $A$ corresponds to $P$ and $B$ corresponds to $Q$. Matching: $A \leftrightarrow P$,$B \leftrightarrow Q$,$C \leftrightarrow R$. Therefore,$\Delta ABC \cong \Delta PQR$.

In $\Delta ABC$ and $\Delta PQR$,if $\frac{AB}{PQ} = \frac{BC}{RQ} = \frac{AC}{PR} = 1$,then $\Delta ABC \cong \Delta \ldots \ldots \ldots$

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