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I'm trying to understand the following passage from Boyer's and Merzbach's History of Mathematics:

page 203

Fig. 9.1

The question I have is: how does the author derive that $w^2=2rv$?

Adrien Hingert
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    $w^2 = v^2 + \frac{s^2}{4}$, $v=r-u$, and $u^2 = r^2 - \frac{s^2}{4}$. Substituting $v$ and $\frac{s^2}{4}$ in the first equation from information in the latter two equations gives us $w^2 = (r-u)^2 + r^2 - u^2 = 2r^2 - 2ru =2r(r-u)=2rv$ so $w^2 = 2rv$ – Sam Mar 22 '23 at 06:13
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    @Sam That's a full answer. Why have you put it in the comment section? – Arthur Mar 22 '23 at 06:14
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    @Arthur It felt wrong to put such a short answer as a proper answer – Sam Mar 22 '23 at 06:19
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    @Sam you can also post as "Community Wiki" if you don't want the reputation for it, but at least the question gets an answer that can be accepted. – b00n heT Mar 22 '23 at 07:33
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    @Sam If the answer to a question is a simple calculation, then the answer to a question is a simple calculation. It still belongs in the answer section. Because it's an answer. – Arthur Mar 22 '23 at 08:27

1 Answers1

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Typing out Sam's solution from the comment section:

$w^2=v^2+\frac{s^2}4$, $v=r−u$, and $u^2=r^2−\frac{s^2}4$. Substituting $v$ and $\frac{s^2}4$ in the first equation from information in the latter two equations gives us $$w^2=(r−u)^2+r^2−u^2=2r^2−2ru=2(r−u)=2rv$$ so $w^2=2rv$.

Arthur
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