Question:
If in $\triangle A B C$ and $\Delta P Q R$, we have: $\frac{A B}{Q R}=\frac{B C}{P R}=\frac{C A}{P Q}$, then
(a) ∆PQR ∼ ∆CAB
(b) ∆PQR ∼ ∆ABC
(c) ∆CBA ∼ ∆PQR
(d) ∆BCA ∼ ∆PQR
Solution:
(a) ∆PQR ∼ ∆CAB
In ∆ABC and ∆PQR, we have:
$\frac{A B}{Q R}=\frac{B C}{P R}=\frac{C A}{P Q}$
$\Rightarrow \triangle A B C \sim \triangle Q R P$
We can also write it as $\triangle P Q R \sim \triangle C A B$.
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