The following triangle ABC, has a right-angle at B. The line BX is perpendicular to AC. The
angle ABX is α and the angle at XBC is 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] ■
[1.1]
[1.2]
[1.3]