Proof: Pythagorean Theorem
Let's prove the following theorem:
if ∠ZXY is a right angle, then ((distance XY) ⋅ (distance XY)) + ((distance XZ) ⋅ (distance XZ)) = (distance YZ) ⋅ (distance YZ)
Proof:
Given
Additional Assumptions
1 | ∠ZXY is a right angle |
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2 | ∠XPY is a right angle |
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3 | m∠YPZ = 180 |
# | Claim | Reason |
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1 | △PYX ∼ △XYZ | if ∠ZXY is a right angle and ∠XPY is a right angle and m∠YPZ = 180, then △PYX ∼ △XYZ |
2 | △PXZ ∼ △XYZ | if ∠ZXY is a right angle and ∠XPY is a right angle and m∠YPZ = 180, then △PXZ ∼ △XYZ |
3 | (distance XY) / (distance ZY) = (distance YP) / (distance YX) | if △PYX ∼ △XYZ, then (distance XY) / (distance ZY) = (distance YP) / (distance YX) |
4 | (distance XZ) / (distance YZ) = (distance ZP) / (distance ZX) | if △PXZ ∼ △XYZ, then (distance XZ) / (distance YZ) = (distance ZP) / (distance ZX) |
5 | ((distance XY) ⋅ (distance XY)) + ((distance XZ) ⋅ (distance XZ)) = ((distance YZ) ⋅ (distance YP)) + ((distance YZ) ⋅ (distance ZP)) | if (distance XY) / (distance ZY) = (distance YP) / (distance YX) and (distance XZ) / (distance YZ) = (distance ZP) / (distance ZX), then ((distance XY) ⋅ (distance XY)) + ((distance XZ) ⋅ (distance XZ)) = ((distance YZ) ⋅ (distance YP)) + ((distance YZ) ⋅ (distance ZP)) |
6 | ((distance YZ) ⋅ (distance YP)) + ((distance YZ) ⋅ (distance ZP)) = (distance YZ) ⋅ ((distance YP) + (distance ZP)) | ((distance YZ) ⋅ (distance YP)) + ((distance YZ) ⋅ (distance ZP)) = (distance YZ) ⋅ ((distance YP) + (distance ZP)) |
7 | (distance YP) + (distance ZP) = distance YZ | if m∠YPZ = 180, then (distance YP) + (distance ZP) = distance YZ |
8 | ((distance YZ) ⋅ (distance YP)) + ((distance YZ) ⋅ (distance ZP)) = (distance YZ) ⋅ (distance YZ) | if ((distance YZ) ⋅ (distance YP)) + ((distance YZ) ⋅ (distance ZP)) = (distance YZ) ⋅ ((distance YP) + (distance ZP)) and (distance YP) + (distance ZP) = distance YZ, then ((distance YZ) ⋅ (distance YP)) + ((distance YZ) ⋅ (distance ZP)) = (distance YZ) ⋅ (distance YZ) |
9 | ((distance XY) ⋅ (distance XY)) + ((distance XZ) ⋅ (distance XZ)) = (distance YZ) ⋅ (distance YZ) | if ((distance XY) ⋅ (distance XY)) + ((distance XZ) ⋅ (distance XZ)) = ((distance YZ) ⋅ (distance YP)) + ((distance YZ) ⋅ (distance ZP)) and ((distance YZ) ⋅ (distance YP)) + ((distance YZ) ⋅ (distance ZP)) = (distance YZ) ⋅ (distance YZ), then ((distance XY) ⋅ (distance XY)) + ((distance XZ) ⋅ (distance XZ)) = (distance YZ) ⋅ (distance YZ) |
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