# If in ∆ABC and ∆PQR, we have:

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 ∆PQRwe 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 $\triangle P Q R \sim \triangle C A B$.