Question Number 208455 by efronzo1 last updated on 16/Jun/24
Commented by mr W last updated on 16/Jun/24
$${the}\:{green}\:{one}\:{even}\:{doesn}'{t}\:{need}\:{to}\:{be}\: \\ $$$${a}\:{semicircle}. \\ $$
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
$$\frac{{b}+{c}}{{z}}=\frac{{z}}{{b}}\:\Rightarrow{b}\left({b}+{c}\right)={z}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} ={b}\left({b}+{c}\right)={z}^{\mathrm{2}} \\ $$$$\Rightarrow{a}={z}\:\checkmark \\ $$