Proof: Similar Corresponding Medians

Let's prove the following theorem:

if △ABC ∼ △XYZ and M is the midpoint of line AC and N is the midpoint of line XZ, then (distance BM) / (distance YN) = (distance BC) / (distance YZ)

Proof:

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Given

1	△ABC ∼ △XYZ
2	M is the midpoint of line AC
3	N is the midpoint of line XZ

Proof Table

#	Claim	Reason
1	m∠BCA = m∠YZX	if △ABC ∼ △XYZ, then m∠BCA = m∠YZX (Similar Triangles Property)
2	(distance CA) / (distance ZX) = (distance BC) / (distance YZ)	if △ABC ∼ △XYZ, then (distance CA) / (distance ZX) = (distance BC) / (distance YZ) (Proportions in Similar Triangles 2)
3	(distance CM) ⋅ 2 = distance CA	if M is the midpoint of line AC, then (distance CM) ⋅ 2 = distance CA (Midpoint Distance 2c)
4	(distance ZN) ⋅ 2 = distance ZX	if N is the midpoint of line XZ, then (distance ZN) ⋅ 2 = distance ZX (Midpoint Distance 2c)
5	((distance CM) ⋅ 2) / ((distance ZN) ⋅ 2) = (distance CA) / (distance ZX)	if (distance CM) ⋅ 2 = distance CA and (distance ZN) ⋅ 2 = distance ZX, then ((distance CM) ⋅ 2) / ((distance ZN) ⋅ 2) = (distance CA) / (distance ZX) (Divide Substitute 2)
6	((distance CM) ⋅ 2) / ((distance ZN) ⋅ 2) = (distance CM) / (distance ZN)	((distance CM) ⋅ 2) / ((distance ZN) ⋅ 2) = (distance CM) / (distance ZN) (divide_by_2)
7	(distance CM) / (distance ZN) = (distance CA) / (distance ZX)	if ((distance CM) ⋅ 2) / ((distance ZN) ⋅ 2) = (distance CM) / (distance ZN) and ((distance CM) ⋅ 2) / ((distance ZN) ⋅ 2) = (distance CA) / (distance ZX), then (distance CM) / (distance ZN) = (distance CA) / (distance ZX) (Transitive Property of Equality Variation 2)
8	(distance CM) / (distance ZN) = (distance BC) / (distance YZ)	if (distance CM) / (distance ZN) = (distance CA) / (distance ZX) and (distance CA) / (distance ZX) = (distance BC) / (distance YZ), then (distance CM) / (distance ZN) = (distance BC) / (distance YZ) (Transitive Property of Equality)
9	(distance BC) / (distance YZ) = (distance CM) / (distance ZN)	if (distance CM) / (distance ZN) = (distance BC) / (distance YZ), then (distance BC) / (distance YZ) = (distance CM) / (distance ZN) (Symmetric Property of Equality)
10	m∠AMC = 180	if M is the midpoint of line AC, then m∠AMC = 180 (Angle Formed by Midpoint)
11	m∠XNZ = 180	if N is the midpoint of line XZ, then m∠XNZ = 180 (Angle Formed by Midpoint)
12	m∠BCM = m∠YZN	if m∠AMC = 180 and m∠XNZ = 180 and m∠BCA = m∠YZX, then m∠BCM = m∠YZN (Two Collinear Angles)
13	△BCM ∼ △YZN	if m∠BCM = m∠YZN and (distance BC) / (distance YZ) = (distance CM) / (distance ZN), then △BCM ∼ △YZN (If Sas Then Similar Triangles)
14	(distance BM) / (distance YN) = (distance BC) / (distance YZ)	if △BCM ∼ △YZN, then (distance BM) / (distance YN) = (distance BC) / (distance YZ) (Proportions in Similar Triangles 3)

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