the ASA Postulate
In the given picture , we have two triangles ∆LKM and ∆JKM , in which we have
By using ASA congruence postulate , we have
∆LKM and ∆JKM
ASA congruence postulate tells that if two angles and the included side of a triangle are congruent to two angles and the included side of other triangle then the triangles are congruent.
The HL Theorem.
Step-by-step explanation: We are given that the triangles have a congruent 90 degree angle and two congruent sides. This would indicate towards SAS but in this case, that doesn't work because the angle is not in between the two congruent sides. We know that the triangles are right, and that their hypotenuses are congruent and that one leg is congruent. Thus, the answer must be HL (hypotenuse & leg).
d.) the HL theorem
It's SAS POSTULATE
Yes, you have chosen the correct answer, it should be SAS postulate, because from figure, we know that we have 2 sides of triangles and its one angle as equal.
So it would be efficient to use SAS postulate, as we have parameters to prove this true using this postulate.
The HL theorem ⇒ answer (b)
* 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)
trust my last answer