Answer:
A: constructing ΔABC so that AB ≅ XY, BC ≅ YZ, and CA ≅ ZX
C: constructing ΔABC so that AB ≅ XY, CA ≅ ZX, and ∠A ≅ ∠X
Step-by-step explanation:
* Lets revise the ways of constructing a triangle
- There are 3 ways to construct a triangle
# You must have length of one side and measures of two angles which
are the endpoint of the side
# You must have a length of two sides and the measure of the including
angle between the two sides
# You must have the length of the 3 sides
* Lets solve the problem
∵ Δ XYZ ≅ Δ ABC
∴ XY = AB , YZ = BC , ZX = CA
∴ m∠ X = m∠ A , m∠ Y = m∠ B , m∠ Z = m∠ C
- To construct Δ ABC You can use:
# The length of the 3 sides
∵ AB ≅ XY , BC ≅ YZ , CA ≅ ZX
∴ constructing ΔABC so that AB ≅ XY, BC ≅ YZ, and CA ≅ ZX ⇒ A
# The length of two sides and the measure of the including angle
∵ AB ≅ XY , CA = ZX
∵ The including angles between the sides are ∠A and ∠X
∴ constructing ΔABC so that AB ≅ XY, CA ≅ ZX, and ∠A ≅ ∠X ⇒ C
# The length of one side and the measures of two angles which are
the endpoints of the side
∵ CA ≅ ZX
∵ The end points are C , A and Z , X
∴ constructing ΔABC so that CA ≅ ZX, ∠A ≅ ∠X, and ∠C ≅ ∠Z
- But this answer not in the choices
∴ The answer is A and C
by a linear divisor 
is simply 
.
, then the remainder upon dividing by 
is