Proof: Similar Distances 2
Let's prove the following theorem:
if △ABC ∼ △XYZ, then (distance CB) / (distance ZY) = (distance BA) / (distance YX)
Proof:
Given
1 | △ABC ∼ △XYZ |
---|
# | Claim | Reason |
---|---|---|
1 | (distance CB) / (distance ZY) = (distance CA) / (distance ZX) | if △ABC ∼ △XYZ, then (distance CB) / (distance ZY) = (distance CA) / (distance ZX) |
2 | (distance CA) / (distance ZX) = (distance BA) / (distance YX) | if △ABC ∼ △XYZ, then (distance CA) / (distance ZX) = (distance BA) / (distance YX) |
3 | (distance CB) / (distance ZY) = (distance BA) / (distance YX) | if (distance CB) / (distance ZY) = (distance CA) / (distance ZX) and (distance CA) / (distance ZX) = (distance BA) / (distance YX), then (distance CB) / (distance ZY) = (distance BA) / (distance YX) |
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