Question Number 207597 by hardmath last updated on 19/May/24
Answered by A5T last updated on 20/May/24
Commented by A5T last updated on 20/May/24
$$\frac{{a}}{{x}}=\frac{{b}}{{y}}\Rightarrow\frac{{a}}{{b}}=\frac{{x}}{{y}}\Rightarrow\frac{{x}}{{x}+{y}}=\frac{{a}}{{a}+{b}} \\ $$$$\frac{{z}}{{b}}=\frac{{a}}{{a}+{b}}\Rightarrow{z}=\frac{{ab}}{{a}+{b}} \\ $$$$\frac{{d}}{{a}}=\frac{\frac{{ab}}{{a}+{b}}}{{b}}=\frac{{a}}{{a}+{b}}\Rightarrow{a}−{d}=\frac{−{ab}}{{a}+{b}} \\ $$$${c}^{\mathrm{2}} =\left({a}−{d}\right)^{\mathrm{2}} +{z}^{\mathrm{2}} =\frac{\mathrm{2}\left({ab}\right)^{\mathrm{2}} }{\left({a}+{b}\right)^{\mathrm{2}} }\Rightarrow{c}=\frac{{ab}\sqrt{\mathrm{2}}}{{a}+{b}} \\ $$
Commented by hardmath last updated on 20/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$
Answered by mr W last updated on 20/May/24
$$\frac{\frac{{c}}{\:\sqrt{\mathrm{2}}}}{{b}−\frac{{c}}{\:\sqrt{\mathrm{2}}}}=\frac{{a}}{{b}}\:\Rightarrow\frac{{c}}{\:\sqrt{\mathrm{2}}{b}−{c}}=\frac{{a}}{{b}}\:\Rightarrow{c}=\frac{\sqrt{\mathrm{2}}{ab}}{{a}+{b}}\:\checkmark \\ $$
Commented by hardmath last updated on 20/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{professor} \\ $$
Commented by mr W last updated on 20/May/24
Commented by A5T last updated on 20/May/24
$${Using}\:{areas}:\:\frac{{bc}}{\mathrm{2}\sqrt{\mathrm{2}}}+\frac{{ac}}{\mathrm{2}\sqrt{\mathrm{2}}}=\frac{{ab}}{\mathrm{2}}\Rightarrow{c}=\frac{\sqrt{\mathrm{2}}{ab}}{{a}+{b}} \\ $$
Answered by efronzo1 last updated on 20/May/24
$$\:\:\:\:\cancel{\underline{\lessdot}}\bigtriangleup\mathrm{ABC}\:=\:\:\cancel{\underline{\lessdot}}\bigtriangleup\mathrm{ABD}\:+\:\:\underline{\vdots}\underbrace{\spadesuit\spadesuit\spadesuit\mathscr{H}\:} \\ $$