Let AB and BC be the two given straight lines.
It is required to find a mean proportional to AB and BC.
Place them in a straight line, and describe the semicircle ADC on AC. Draw BD from the point B at right angles to the straight line AC, and join AD and DC.
Since the angle ADC is an angle in a semicircle, it is right.
And, since, in the right-angled triangle ADC, BD has been drawn from the right angle perpendicular to the base, therefore BD is a mean proportional between the segments of the base, AB and BC.
Therefore a mean proportional BD has been found to the two given straight lines AB and BC.
This construction of the mean proportional was used before in II.14 to find a square equal to a given rectangle. By proposition VI.17 coming up, the two constructions are equivalent. That is the mean proportional between two lines is the side of a square equal to the rectangle contained by the two lines. Algebraically, a : x = x : b if and only if ab = x2. Thus, x is the square root of ab. This mean proportional between a and b is also called the geometric mean of a and b.
When b is taken to have unit length, this construction gives the construction for the square root of a.
This construction is used in the proofs of propositions VI.25, X.27, and X.28.
| Book VI | VI.25 |
|---|---|
| Book X | X.27, X.28 |