How did ancient Greeks use geometry to solve algebraic equations?

[Yann]

I've heard ancient greeks were able to solve algebra using geometry.

For example, if you have a line between two points AC (B being right in the middle), can you find a point x so;

AB/AX = AX/AC

if a is length of AB, then AB = a and AC = 2a, so the problem is really about;

a/x = x/a2; 2a^2 = x^2; 2^(1/2)a = x, if a = 1 then x = 2^(1/2).

How could they find the point x ?

 

[robert Ihnot]

Are you talking about the golden ratio? Where the large length is to the smaller segment as the whole line is to the large part? Let x = long part, and L = total length: x/(L-x) =L/x

 

[Hurkyl]

AB/AX = AX/AC

Cross ratios are easy! Note that's the same as:

AB * AC = AX * AX

So, you just make a line segment:

C ----- D ------------ E

with CD = AB and DE = AC. Then you construct a circle with CE as its diameter. Draw the perpendicular at D, which intersects the circle at points F and G. Note that DF and DG are congruent.

Now, you have two chords intersecting, and thus you have:

DC * DE = DF * DG

and so

AB * AC = DF * DF

So DF is the length you seek.