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:
Given
1 | △ABC ∼ △XYZ |
---|---|
2 | M is the midpoint of line AC |
3 | N is the midpoint of line XZ |
# | Claim | Reason |
---|---|---|
1 | m∠BCA = m∠YZX | if △ABC ∼ △XYZ, then m∠BCA = m∠YZX |
2 | (distance CA) / (distance ZX) = (distance BC) / (distance YZ) | if △ABC ∼ △XYZ, then (distance CA) / (distance ZX) = (distance BC) / (distance YZ) |
3 | (distance CM) ⋅ 2 = distance CA | if M is the midpoint of line AC, then (distance CM) ⋅ 2 = distance CA |
4 | (distance ZN) ⋅ 2 = distance ZX | if N is the midpoint of line XZ, then (distance ZN) ⋅ 2 = distance ZX |
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) |
6 | ((distance CM) ⋅ 2) / ((distance ZN) ⋅ 2) = (distance CM) / (distance ZN) | ((distance CM) ⋅ 2) / ((distance ZN) ⋅ 2) = (distance CM) / (distance ZN) |
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) |
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) |
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) |
10 | m∠AMC = 180 | if M is the midpoint of line AC, then m∠AMC = 180 |
11 | m∠XNZ = 180 | if N is the midpoint of line XZ, then m∠XNZ = 180 |
12 | m∠BCM = m∠YZN | if m∠AMC = 180 and m∠XNZ = 180 and m∠BCA = m∠YZX, then m∠BCM = m∠YZN |
13 | △BCM ∼ △YZN | if m∠BCM = m∠YZN and (distance BC) / (distance YZ) = (distance CM) / (distance ZN), then △BCM ∼ △YZN |
14 | (distance BM) / (distance YN) = (distance BC) / (distance YZ) | if △BCM ∼ △YZN, then (distance BM) / (distance YN) = (distance BC) / (distance YZ) |
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