If in two triangles $ABC$ and $PQR$,$\frac{AB}{QR} = \frac{BC}{PR} = \frac{CA}{PQ}$,then

  • A
    $\triangle PQR \sim \triangle ABC$
  • B
    $\triangle PQR \sim \triangle CAB$
  • C
    $\triangle CBA \sim \triangle PQR$
  • D
    $\triangle BCA \sim \triangle PQR$

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Similar Questions

In $\Delta ABC$,$m\angle B = 90^\circ$. If $AB = 12$ and $BC = 5$,then $AC = \dots$

In $\Delta ABC$,$A-P-B$,$A-Q-C$ and $\overline{PQ} \parallel \overline{BC}$. Then,$\ldots \ldots \ldots$ holds good.

In $\square ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{M\}$. If $MA = 12$,$AC = 20$,and $MB = 9$,find $MD$.

$ABCD$ is a rectangle. If $AB = 7.5$ and $AC = 19.5$,find $BC$.

In $\Delta ABC$ and $\Delta PQR$,$m \angle A = 30^{\circ}$,$m \angle C = 60^{\circ}$,$m \angle P = 90^{\circ}$ and $m \angle Q = 30^{\circ}$. Then,the correspondence $ABC \leftrightarrow \ldots$ is a similarity.

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