Question

which would prove that δabc ~ δxyz? select two options.
□ \\(\frac{ba}{yx} = \frac{bc}{yz} = \frac{ac}{xz}\\)
□ \\(\frac{ba}{yx} = \frac{bc}{yz}, \angle c \cong \angle z\\)
□ \\(\frac{ac}{xz} = \frac{ba}{yx}, \angle a \cong \angle x\\)
□ \\(\frac{ba}{yx} = \frac{ac}{yz} = \frac{bc}{xz}\\)
□ \\(\frac{bc}{xy} = \frac{ba}{zx}, \angle c \cong \angle x\\)

Explanation:

Step1: Check SSS similarity

SSS similarity requires all corresponding sides proportional. The first option has $\frac{BA}{YX}=\frac{BC}{YZ}=\frac{AC}{XZ}$, matching SSS.

Step2: Check SAS similarity

SAS similarity needs two proportional sides and included angle congruent. The third option has $\frac{AC}{XZ}=\frac{BA}{YX}$ and $\angle A \cong \angle X$ (included angle), matching SAS.

Step3: Verify other options

Other options fail (e.g., second option has non-included angle; fourth/fifth have mismatched ratios/angles).

Answer:

$\frac{BA}{YX} = \frac{BC}{YZ} = \frac{AC}{XZ}$, $\frac{AC}{XZ} = \frac{BA}{YX}, \angle A \cong \angle X$