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Question Number 61347 by Tawa1 last updated on 01/Jun/19

Commented by mr W last updated on 02/Jun/19

$$x=20°,seealsoQ43728.$$

Answered by peter frank last updated on 01/Jun/19

$$x=40°$$

Answered by mr W last updated on 02/Jun/19

Commented by mr W last updated on 02/Jun/19

$$drawDF//AB$$$$\mathrm{\Delta }DFGisequilateral,\Rightarrow DF=GF$$$$\mathrm{\Delta }AFCisisosceles,\Rightarrow AF=CF$$$$\mathrm{\Delta }AGC\equiv \mathrm{\Delta }CEA,\Rightarrow AG=CE$$$$AF-AG=CF-CE$$$$\Rightarrow GF=EF$$$$\Rightarrow \mathrm{\Delta }DFEisisosceles,$$$$\Rightarrow \mathrm{\angle }DEF=\mathrm{\angle }EDF=\frac{180-\mathrm{\angle }DEF}{2}=\frac{180-80}{2}=50°$$$$\mathrm{\angle }AEB=180-70-80=30°$$$$x=\mathrm{\angle }DEF-\mathrm{\angle }AEB=50-30=20°$$

Commented by otchereabdullai@gmail.com last updated on 11/Apr/20

$$\mathrm{wow}$$

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