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:

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Given

1	∠ZXY is a right angle

Additional Assumptions

2	∠XPY is a right angle
3	m∠YPZ = 180

Proof Table

#	Claim	Reason
1	△PYX ∼ △XYZ	if ∠ZXY is a right angle and ∠XPY is a right angle and m∠YPZ = 180, then △PYX ∼ △XYZ (Similar Triangles Example 2)
2	△PXZ ∼ △XYZ	if ∠ZXY is a right angle and ∠XPY is a right angle and m∠YPZ = 180, then △PXZ ∼ △XYZ (Similar Triangles Example 3)
3	(distance XY) / (distance ZY) = (distance YP) / (distance YX)	if △PYX ∼ △XYZ, then (distance XY) / (distance ZY) = (distance YP) / (distance YX) (Similar Distances 2)
4	(distance XZ) / (distance YZ) = (distance ZP) / (distance ZX)	if △PXZ ∼ △XYZ, then (distance XZ) / (distance YZ) = (distance ZP) / (distance ZX) (Proportions in Similar Triangles 9)
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)) (Distance Algebra Example)
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)) (Distributive Property Variation 4)
7	(distance YP) + (distance ZP) = distance YZ	if m∠YPZ = 180, then (distance YP) + (distance ZP) = distance YZ (Collinear Points Property 2)
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) (Substitution 12)
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) (Transitive Property of Equality)

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