Proof that the distance of the sun from the earth is greater than eighteen times, but less than twenty times, the distance of the moon from the earth:

Let A be the center of the sun, and B that of the earth.

Produce line AB to join these.

Let C be the center of the moon when halved;

let a plane be carried through AB and C, and let the section made by the plane in the sphere on which the center of the sun moves be the great circle ADE.

Produce lines AC and BC, and extend BC to point D.

Then, because the point C is the center of the moon when halved, the angle ACB will be right. (That is, equal to 90 degrees.)

Draw BE from B at right angles to BA;

then the circumference ED will be one-thirtieth of the circumference EDA; for by hypothesis four, when the moon appears to us halved, its distance from the sun is less than a quadrant by one thirtieth of a quadrant.

Thus the angle EBC is also one-thirtieth of a right angle.

Complete the parallelogram AE, and join line BF.

Angle FBE will be half a right angle. (45 degrees)

Let angle FBE be bisected by the straight line BG; therefore the angle GBE is one fourth part of a right angle. (22.5 degrees)

But the angle DBE is also one thirtieth part of a right angle; therefore the ratio of angle GBE to angle DBE is 15 to 2:

for if a right angle is regarded as divided into 60 equal parts, the angle GBE contains 15 of such parts, and the angle DBE contains 2.

Now, since GE has to EH a ratio greater than that which the angle GBE has to angle DBE, GE has to EH a ratio greater than 15 to 2.

Next, since BE=EF, and angle E is 90 degrees, the square of FB is double the square of BE.

But, as the square of FB is to the square of BE, so is the square of FG to the square of GE; therefore FG squared is double GE squared.

Now, 49 is less than double of 25, so the ratio FG squared to GE squared is greater than 49 to 25; therefore the ratio FG to GE is greater than 7 to 5.

Similarly, the ratio FE to EG is greater than 12 to 5, thus is greater then 36 to 15.

But it was also proved that the ratio GE to EH is greater than 15 to 2; therefore by equal measures, the ratio FE to EH is greater than 36 to 2, thus greater than 18 to 1; therefore FE is greater than 18 times EH.

FE=BE; therefore BE is also greater than 18 times EH; therefore BH is much greater than 18 times HE.

But, as BH is to HE, so is AB to BC, because of similar triangles; therefore AB is also greater than 18 times BC.

AB is the distance of the sun from the earth, while CB is the distance of the moon from the earth; therefore the distance of the sun from the earth is greater than 18 times the distance of the moon from the earth.

Again, I say that it is also less than 20 times that distance.

For if DK is drawn through D parallel to EB, and circle DKB is drawn about the triangle DKB, then DB will be its diameter because the angle at K is 90 degrees.

Let BL, the side of a hexagon, be fitted into the circle.

Then, since the angle DBE is 1/30th of a right angle (3 degrees) the angle BDK is also 3 degrees; therefore the circumference BK is 1/60th of the whole circle.

But BL is 1/6th part of the whole circle.

Therefore the circumference BL is ten times circumference BK.

And the ratio of circumferences BL to BK is greater than the ratio of the line segments BL to BK; therefore the straight line BL is less than ten times the straight line BK.

And BD is double of BL; therefore BD is less than 20 times BK.

But, as BD is to BK, so is AB to BC; therefore AB is also less than 20 times BC.

And AB is the distance of the sun from the earth, while BC is the distance of the moon from the earth; therefore the distance of the sun from the earth is less than 20 times the distance of the moon from the earth.

And it was proved before that it is greater than 18 times that distance.