Question Number 195287 by Mingma last updated on 29/Jul/23
Answered by MM42 last updated on 29/Jul/23
$${CD}={DE}\Rightarrow\angle{C}\mathrm{2}=\angle{E}\:\:\:\&\:\angle{D}\mathrm{1}+\angle{E}=\mathrm{60} \\ $$$$\angle{C}\mathrm{1}=\mathrm{60}−\angle{C}\mathrm{2}=\mathrm{60}−\angle{E}=\angle{D}\mathrm{1} \\ $$$$\frac{{CD}}{{Sin}\mathrm{60}}=\frac{\mathrm{2}}{{SinC}\mathrm{1}}\:\:\&\:\:\frac{{DE}}{{Sin}\mathrm{120}}=\frac{{BE}}{{SinD}\mathrm{1}} \\ $$$$\frac{\mathrm{2}}{{SinC}\mathrm{1}}=\frac{{BE}}{{SinD}\mathrm{1}}\Rightarrow{BE}=\mathrm{2}\:\checkmark \\ $$
Answered by ajfour last updated on 29/Jul/23
$${side}\:{AB}={s} \\ $$$${AD}={a}\:\:,\:{BE}={x}=? \\ $$$${CD}^{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{4}}{s}^{\mathrm{2}} +\left(\frac{{s}}{\mathrm{2}}−{a}\right)^{\mathrm{2}} \\ $$$$\left(\frac{{s}+{x}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\left({s}−{a}\right)^{\mathrm{2}} ={DE}^{\:\mathrm{2}} \\ $$$${since}\:\:{DE}={CD} \\ $$$${s}^{\mathrm{2}} −{as}+{a}^{\mathrm{2}} ={s}^{\mathrm{2}} +\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{{sx}}{\mathrm{2}}−\frac{\mathrm{3}{as}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4}}{a}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{{sx}}{\mathrm{2}}=\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+\frac{{sa}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\left({x}+{s}\right)^{\mathrm{2}} =\left({a}+{s}\right)^{\mathrm{2}} \\ $$$$\:{x}={a}\:\:\:\left({here}\:{a}=\mathrm{2}\right) \\ $$