Proof: Similar Triangles Theorem
Let's prove the following theorem:
if m∠CAB = m∠ZXY and m∠ABC = m∠XYZ, then △ABC ∼ △XYZ
Proof:
Given
Additional Assumptions
| 1 | m∠CAB = m∠ZXY |
|---|---|
| 2 | m∠ABC = m∠XYZ |
| 3 | △ABC ∼ △LMN |
|---|---|
| 4 | distance LM = distance XY |
| # | Claim | Reason |
|---|---|---|
| 1 | m∠CAB = m∠NLM | if △ABC ∼ △LMN, then m∠CAB = m∠NLM |
| 2 | m∠ABC = m∠LMN | if △ABC ∼ △LMN, then m∠ABC = m∠LMN |
| 3 | m∠NLM = m∠ZXY | if m∠CAB = m∠NLM and m∠CAB = m∠ZXY, then m∠NLM = m∠ZXY |
| 4 | m∠LMN = m∠XYZ | if m∠ABC = m∠LMN and m∠ABC = m∠XYZ, then m∠LMN = m∠XYZ |
| 5 | △NLM ≅ △ZXY | if m∠NLM = m∠ZXY and distance LM = distance XY and m∠LMN = m∠XYZ, then △NLM ≅ △ZXY |
| 6 | △LMN ≅ △XYZ | if △NLM ≅ △ZXY, then △LMN ≅ △XYZ |
| 7 | △LMN ∼ △XYZ | if △LMN ≅ △XYZ, then △LMN ∼ △XYZ |
| 8 | △ABC ∼ △XYZ | if △ABC ∼ △LMN and △LMN ∼ △XYZ, then △ABC ∼ △XYZ |
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