Commentary on Proposition V.21¶

Algebraically, if

$$\begin{aligned} & a : b = e : f, \\ \text{and} \quad & b : c = d : e, \end{aligned}$$

$$\text{then, according as } a > = < c, \quad d > = < f.$$

Simson’s alterations correspond to those which he makes in Prop. 20. After the first case he proceeds thus.

“Secondly, let A be equal to C; then shall D be equal to F. Because A and C are equal,

[V. 7]A is to B as C is to B.

But A is to B as E is to F, and C is to B as E is to D:

[V. 11]wherefore E is to F as E to D,

[V. 9]and therefore D is equal to F.

Next, let A be less than C; then shall D be less than F. For C is greater than A, and, as was shown,

C is to B as E to D,

and, in like manner,

B is to A as F to E;

therefore F is greater than D, by the first case, and therefore D is less than F.”

The proof may be shown thus.

$$\text{According as } a > = < c, \quad a : b > = < c : b.$$

But $a : b = e : f$, and, by inversion, $c : b = e : d$.

$$\begin{aligned} \text{Therefore, according as } a > = < c, \quad & e : f > = < e : d, \\ \text{and therefore } & d > = < f. \end{aligned}$$