Question

The proof ABC ≅ DCB that is shown.

Given: A ≅ D; CD||AB
Prove: ABC ≅ DCB


What is the missing reason in the proof?



alt. ext. s are ≅
ASA
AAS
corr. int. s are ≅

Answer 1

Given: ∠A≅∠D and CD║AB

To Prove: Δ ABC ≅ ΔD CB

Proof:

1. ∠A=∠D→→(Given)

2. As, CD ║AB, ⇒ ∠DCB=∠ABC→→(when lines are parallel, Alternate interior angles are equal.)

3. Segment CB=Segment BC→→Reflection property

⇒Δ ABC ≅ ΔD CB →→(Angle Angle Side or AAS)

Answer 2

The correct answer is:

AAS

Step-by-step explanation:

AAS stands for "angle-angle-side." This states that if two angles and a non-included side of one triangle are congruent to the corresponding two angles and non-included side of another triangle, then the triangles are congruent.

In these triangles, we have ∠CAB ≅ ∠CDB given to us to begin with. Throughout the proof, we find that ∠ABC ≅ ∠DCB. We also have that CB is congruent to itself. This is two angles and a side not included, so this is AAS.