Similar Distances 2

Proof: Similar Distances 2

Let's prove the following theorem:

if △ABC ∼ △XYZ, then (distance CB) / (distance ZY) = (distance BA) / (distance YX)

Proof:

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Given

1	△ABC ∼ △XYZ

Proof Table

#	Claim	Reason
1	(distance CB) / (distance ZY) = (distance CA) / (distance ZX)	if △ABC ∼ △XYZ, then (distance CB) / (distance ZY) = (distance CA) / (distance ZX) (Proportions in Similar Triangles)
2	(distance CA) / (distance ZX) = (distance BA) / (distance YX)	if △ABC ∼ △XYZ, then (distance CA) / (distance ZX) = (distance BA) / (distance YX) (Proportions in Similar Triangles 8)
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) (Transitive Property of Equality)

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