Question

Proving Pythagoras...

Answer 1

Right Angles:
CEO
BCA
ODC
CFO

Area of ∆BCO = CB*EO/2
tan(x/2) = (a/2)/n
n = a/2tan(x/2)

let t = tan(x/2)
n = a/2t
Area of ∆BCO = a^2/4t

similarly Area of ∆ACO = b^2/4T = tb^2/4
where T = tan((180-x)/2) = 1/t

sin(x) = h/r
h = r*sin(x)

Area of ∆ABC = 2r*r*sin(x)/2 = r^2*sin(x)

c = 2r
r = c/2

Area of ∆ABC = c^2*sin(x)/4 = c^2*2t/(4+4t^2) = tc^2/(2+2t^2)

a^2/4t+tb^2/4/2=tc^2/(2+2t^2)

we know that a^2 + b^2 = c^2

can we show this from what we have?
Also, are there any errors in my work?

Answer 2

^actually, a^2/4t+tb^2/4=tc^2/(2+2t^2)

Answer 3

nice work so far.. im also not sure how to take it from here. id love to see if you can prove it.... good luck :)

Answer 4

Sure Thanks