Commentary on Proposition VIII.5

Back to Proposition VIII.5.

[Folio 0360 | p. 354]

If

$$a = cd,\quad b = ef,$$

then $a$ has to $b$ the ratio compounded of $c : e$ and $d : f$.

Take three numbers the least which are continuously in the given ratios.

[Folio 0361 | p. 355]

If $l$ is the L.C.M. of $e$, $d$ and $l = me = nd$, the three numbers are

$$\left. \begin{array}{r} mc,\, me \\ = nd \end{array} \right\},\, nf.\hfill\text{[VIII. 4]}$$

Now

$$\begin{aligned} dc : de &= c : e\hfill\text{[VII. 17]}\\ &= mc : me = mc : nd. \end{aligned}$$

Also

$$\begin{aligned} ed : ef &= d : f\hfill\text{[VII. 17]}\\ &= nd : nf. \end{aligned}$$

Therefore, ex aequali,

$$\begin{aligned} cd : ef &= mc : nf\\ &= (\text{ratio compounded of } c : e \text{ and } d : f). \end{aligned}$$

It will be seen that this proof follows exactly the method of VI. 23 for parallelograms.