Commentary on Proposition VII.7¶

If $a = \frac{1}{n}b$ and $c = \frac{1}{n}d$, we are to prove that

$$a - c = \frac{1}{n}(b - d),$$

a result differing from that of VII. 5 in that minus is substituted for plus. Euclid’s method is as follows.

Suppose that $e$ is taken such that

$$a - c = \frac{1}{n}e. \tag{1}$$

Now

$$c = \frac{1}{n}d.$$

Therefore

$$a = \frac{1}{n}(d + e), \tag{VII. 5}$$

whence, from the hypothesis, $d + e = b$, so that $e = b - d$, and, substituting this value of $e$ in (1), we have

$$a - c = \frac{1}{n}(b - d).$$