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pochemuha
3 years ago
10

The proof ABC ≅ DCB that is shown.

Mathematics
2 answers:
miskamm [114]3 years ago
8 0

Solution:

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)

Schach [20]3 years ago
4 0

The <em><u>correct answer</u></em> is:

AAS

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.

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