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Question Number 195287 by Mingma last updated on 29/Jul/23

Answered by MM42 last updated on 29/Jul/23

$$CD=DE\Rightarrow \mathrm{\angle }C2=\mathrm{\angle }E\mathrm{\&}\mathrm{\angle }D1+\mathrm{\angle }E=60$$$$\mathrm{\angle }C1=60-\mathrm{\angle }C2=60-\mathrm{\angle }E=\mathrm{\angle }D1$$$$\frac{CD}{Sin60}=\frac{2}{SinC1}\mathrm{\&}\frac{DE}{Sin120}=\frac{BE}{SinD1}$$$$\frac{2}{SinC1}=\frac{BE}{SinD1}\Rightarrow BE=2✓$$

Answered by ajfour last updated on 29/Jul/23

$$sideAB=s$$$$AD=a,BE=x=?$$$${CD}^2=\frac{3}{4}{s}^2+{(\frac{s}{2}-a)}^2$$$${\left(\frac{s+x}{2}\right)}^2+\frac{3}{4}{(s-a)}^2={DE}^2$$$$sinceDE=CD$$$$s^2-as+a^2=s^2+\frac{x^2}{4}+\frac{sx}{2}-\frac{3as}{2}+\frac{3}{4}{a}^2$$$$\Rightarrow \frac{x^2}{4}+\frac{sx}{2}=\frac{a^2}{4}+\frac{sa}{2}$$$$\Rightarrow {(x+s)}^2={(a+s)}^2$$$$x=a(herea=2)$$

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