Question

(1) In ∆ABC and ∆PQR to one correspondence AB/ QR = BC/PR = CA/PQ then.

(1) ∆PQR ~ ∆ABC
(2) ∆PQR~∆CAB
(3)∆CBA~∆PQR
(4)∆BCA~∆PQR​

Answer 1

$\orange{\boxed{\purple{\underline{\red{\mathrm{Solution:-}}}}}}$

$\boxed{Given:-}$

$\frac{AB}{QR}$=$\frac{BC}{PR}$=$\frac{CA}{PQ}$

$\boxed{Proof:-}$

→Since The Corresponding sides are equal we have to check whether while naming the Corresponding sides of the ∆les are equal or not↓

In Option 1,3&4

→we see it doesnt Fulfill the requirements which are given in the Question.

=> Option (2) is correct as it Proves the Requirement of Naming,

→∆PQR≈∆CAB [°≈ this sign = Similar ∆les]

We get,

→$\frac{PQ}{CA}$=$\frac{QR}{AB}$=$\frac{PR}{BC}$

↑its just the opposite of Given

$\mathcal{BE \: BRAINLY}$

Answer 2

$\huge \boxed{ \fcolorbox{cyan}{red}{Answer : }}$

Given:

$\sf{\frac{AB}{QR}} = \frac{BC}{PR} = \frac{CA}{PQ}$

$\rm{option \: 2 \: is \: correct}$

∆PQR≈∆CAB

$\sf{ \frac{PQ}{CA}} = \frac{QR}{AB} = \frac{PR}{BC}$

it's the opposition..

so option 2 is correct

➡ The Corresponding sides are equal we have to check whether while naming the Corresponding sides of the triangles are equal or not