Answer:
The HL theorem ⇒ answer (b)
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
* In the problem
- In Δ ABC
∵ AC = 10 units
∵ BC = 6 units
∵ m∠ABC = 90°
- In Δ DEF
∵ DF = 10 units
∵ EF = 6 units
∵ m∠DEF = 90°
∴ AC = DF = 10 units
∵ BC = EF = 6 units
∵ m∠∠B = m∠E = 90°
∴ The two triangles ABC and DEF are congruent by using
the HL theorem (hypotenuse leg of a right triangle)
