Proof: Pythagorean Theorem

Let's prove the following theorem:

First, we assume that we can draw a perpendicular line from X to YZ.

Notice that PYX and XYZ share the angle PYX. We use this fact to show that △PYX and △XYZ are similar triangles.

Similarly, PXZ and XYZ share the angle PZX. Thus, we can show that △PXZ and △XYZ are also similar triangles.

Then:

$\frac{\mathrm{XY}}{\mathrm{ZY}}=\frac{\mathrm{YP}}{\mathrm{YX}}$

Where XY stands for "distance between X and Y."

We also claim that:

$\frac{\mathrm{XZ}}{\mathrm{YZ}}=\frac{\mathrm{PZ}}{\mathrm{XZ}}$

Then using the cross multiplication theorem, we make the following claim:

XY ⋅ XY = YP ⋅ YZ

XZ ⋅ XZ = PZ ⋅ YZ

We can add the left sides and the right sides to conclude that:

XY ⋅ XY + XZ ⋅ XZ = YP ⋅ YZ + PZ ⋅ YZ

Using the distributive property, we claim that:

YP ⋅ YZ + PZ ⋅ YZ = (YP + PZ) ⋅ YZ

Since YP + PZ = YZ:

YP ⋅ YZ + PZ ⋅ YZ = (YZ) ⋅ YZ

Finally, using the transitive property, we claim that:

XY ⋅ XY + XZ ⋅ XZ = YZ ⋅ YZ

Proof:

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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