Answer:
The answer is i and iii only ⇒ answer C
Step-by-step explanation:
* Lets revise the cases of congruence
- SSS ⇒ 3 sides in the 1st Δ ≡ 3 sides in the 2nd Δ
- SAS ⇒ 2 sides and including angle in the 1st Δ ≡ 2 sides and
including angle in the 2nd Δ
- ASA ⇒ 2 angles and the side whose joining them in the 1st Δ
≡ 2 angles and the side whose joining them in the 2nd Δ
- AAS ⇒ 2 angles and one side in the first triangle ≡ 2 angles
and one side in the 2ndΔ
- HL ⇒ hypotenuse leg of the first right angle triangle ≡ hypotenuse
leg of the 2nd right angle Δ
* Lets solve the problem
- From the figure in Δ ABC and Δ EFG
∵ AB = 2 units
∵ ∠A and ∠B are the end vertex of AB
∵ EF = 2 units
∵∠E and ∠F are the end vertex of EF
∴ Δ ABC is congruent to Δ EFG if:
# m∠B = m∠F
# m∠A = m∠E
∴ The necessary statements to prove that Δ ABC is congruent to
Δ EFG by ASA are:
i. m∠B = m∠F
iii. m∠A = m∠E
* The answer is i and iii only
