At any rate, the method used by Archimedes differs from earlier approximations in a fundamental way. Earlier schemes for approximating pi simply gave an approximate value, usually based on comparing the area or perimeter of a certain polygon with that of a circle. Archimedes' method is new in that it is an iterative process, whereby one can get as accurate an approximation as desired by repeating the process, using the previous estimate of pi to obtain a new one. This is a new feature of Greek mathematics, although it has an ancient tradition among the Chinese in their methods for approximating square roots.
Archimedes' method, as he did it originally, skips over a lot of computational
steps, and is not fully explained, so authors of history of math books
have often presented slight variations on his method to make it easier
to follow. Here we will try to stick to the original as much as possible,
following essentially Heath's translation^{3}.
If an angle of a triangle be bisected and the straight line cutting the angle cut the base also, the segments of the base will have the same ratio as the remaining sides of the triangle; and, if the segments of the base have the same ratio as the remaining sides of the triangle, the straight line joined from the vertex to the point of section will bisect the angle of the triangle.^{4}
Animated GIF Proof
of
Theorem (99K) 

The ratio of the circumference of any circle to its diameter is less than 3^{1}/_{7} but greater than 3^{10}/_{71}.The proof we give below essentially follows that of Archimedes, as set out in Heath's translation^{5}. Much of the text skips over steps in the proof; rather than adding intermediate steps as Heath does^{6}, we are putting those in popup windows. Look for buttons like this: . Clicking on these will bring up popup windows showing intermediate steps that Archimedes has left out of this text (HTGT stands for How'd They Get That?).
I. Let AB be the diameter of any circle, O its center, AC the tangent at A; and let the angle AOC be onethird of a right angle.
and
(2) OC : AC = 306 : 153.
First, draw OD bisecting the angle AOC
and meeting AC in D.
Now
so that
Therefore
Hence
OD^{2} : AD^{2} > 349450 : 23409
so that
(4) OD : DA > 591^{1}/_{8} : 153.
Secondly, let OE bisect the angle AOD,
meeting AD in E.
Therefore
(5) OA : AE > 1162^{1}/_{8} : 153
Thus
(6) OE : EA > 1172^{1}/_{8} : 153.
Thirdly, let OF bisect the angle AOE
and meet AE in F.
We thus obtain the result that
(7) OA : AF > 2334 ^{1}/_{4} : 153
Thus
(8) OF : FA > 2339 ^{1}/_{4} : 153.
Fourthly, let OG bisect the angle AOF,
meeting AF in G.
We have then
OA : AG > 4673 ^{1}/_{2} : 153.
Now the angle AOC, which is onethird of a
right angle, has been bisected four times, and it follows that angle AOG
= 1/48 (a right angle).
Make the angle AOH on the other side of OA
equal to the angle AOG, and let GA produced meet OH in
H.
Then angle GOH = 1/24 (a right angle).
Thus GH is one side of a regular polygon
of 96 sides circumscribed to the given circle.
And, since
OA : AG > 4673 ^{1}/_{2} : 153,
while
AB = 2 OA, GH = 2 AG,
it follows that
AB : (perimeter of a polygon of 96 sides) > 4673 ^{1}/_{2} : 14688
But
Therefore the circumference of the circle (being less than the perimeter
of the polygon) is a fortiori less than 3 1/7 times the diameter
AB.
II. Next let AB be the diameter of a circle, and let AC, meeting the circle in C, make the angle CAB equal to onethird of a right angle. Join BC.
Then
First, let AD bisect the angle BAC
and meet BC in d and the circle in D. Join BD.
Then
angle BAD = angle dAC = angle dBD
and the angles at D, C are both right angles. It follows that the triangles
ADB, BDd are similar.
Therefore
AD : BD = BD : Dd = AB : Bd
= (AB + AC) : (Bd + Cd)
= (AB + AC) : BC
or (BA + AC) : BC = AD : DB.
Therefore
Thus
(2)
AB : BD < 3013 ^{3}/_{4} : 780.
Secondly, let AE bisect the angle
BAD, meeting the circle in E; and let BE be joined.
Then we prove, in the same way as before, that
(3) AE : EB < 5924 ^{3}/_{4} : 780 = 1823 : 240.
Therefore
(4) AB : BE < 1838 ^{9}/_{11} : 240.
Thirdly, let AF bisect the angle BAE, meeting
the circle in F.
Thus,
(5)
AF : FB < 3661 ^{9}/_{11} x ^{11}/_{40}
: 240 x ^{11}/_{40}
= 1007 : 66.
Therefore,
(6) AB : BF < 1009 ^{1}/_{6} : 66.
Fourthly, let the angle BAF be bisected by
AG meeting the circle in G.
Then
AG : GB < 2016 ^{1}/_{6} : 66, by (5) and (6).
Therefore
(7) AB : BG < 2017 ^{1}/_{4} : 66.
Therefore BG is a side of a regular inscribed polygon of 96 sides.
It follows from (7) that
(perimeter of polygon) : AB > 6336 : 2017 ^{1}/_{4}.
And .
Much more then is the circumference to the diameter
< 3 ^{1}/_{7} but >
3 ^{10}/_{71}.