Pythagoras Theorem Proof Using Trigonometry (Similar Triangles)
The
following proof using cosines, and is, therefore, based on similar
triangles. In the ΔABC, below, ∠ABC is a right anlge, and so is ∠BDC.
∠ABC is labelled α, and ∠ACB, β. For the figure, we draw any right
angled triangle and drop a perpendicular from B to D on the line AC. In
the standard way, BC=a, AC=b, and AB=c. We wish to prove for ΔABC, that [1.01] We note that [1.02] Because AC is the sum of the two segments.
From the definition of the cosine, in ΔABD, and ΔADC, and substituting in 1.02 [1.03]
From the definitions of cosine in ΔABC, we have: [1.04] Substituting these values in 1.03: [1.05] And multiplying throughout by b, and simplifying, we have Pythagoras' Theorem. [1.01 repeated]■ Note:
In proving something, we need to guard against assuming the thing we
are proving. The definition of cosine and the claim that it is a
constant is based on similar triangles, and appears not to assume
pythagoras.
Pythagoras Theorem Proof Using Areas
The following triangle ABC, has a right-angle at B. The line BX is perpendicular to AC. The
angle ABX is labelled α and the angle at XBC is labelled β. Because XBC is a right-angled
triangle by construction, then angle C is 90°−angle XBC (β),
so angle C is α. By a similar argument, angle A is β, and these are
marked on the diagram above. The length of BX is h.
It goes like this. The area of triangle ABC is equal to the area of triangle ABX + area of triangle XBC. The area of a triangle can be expressed as: [1.1]
So the areas of the triangles can be expressed: [1.2]
We note that [1.3]
We also note that from the triangle, we can express: cosα=a/b [1.4] cosβ=c/b [1.5] sinα=c/b [1.6] sinβ=a/b [1.7]
Using these in Equation 1.3, we have: h=ca/b [1.8]
Note
also that sin(α+β)=sin (90°)=1, so we can forget sin(α+β) in
Equation 1.2, and substituting for h=ca/b and for the sines:
ca/2=1/2(c·ca/b·c/b+a·ca/b·a/b) [1.9] multiplying throughout by b2/ca, we get Pythagoras' Theorem: b2=a2+c2 [1.10] ■