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Question Number 208455 by efronzo1 last updated on 16/Jun/24

Commented by mr W last updated on 16/Jun/24

the green one even doesn′t need to be a semicircle.

$${the}\:{green}\:{one}\:{even}\:{doesn}'{t}\:{need}\:{to}\:{be}\: \\ $$$${a}\:{semicircle}. \\ $$

Commented by mr W last updated on 16/Jun/24

Answered by mr W last updated on 16/Jun/24

Commented by mr W last updated on 16/Jun/24

((b+c)/z)=(z/b) ⇒b(b+c)=z^2 a^2 =b(b+c)=z^2 ⇒a=z ✓

$$\frac{{b}+{c}}{{z}}=\frac{{z}}{{b}}\:\Rightarrow{b}\left({b}+{c}\right)={z}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} ={b}\left({b}+{c}\right)={z}^{\mathrm{2}} \\ $$$$\Rightarrow{a}={z}\:\checkmark \\ $$

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