Question Number 61347 by Tawa1 last updated on 01/Jun/19
Commented by mr W last updated on 02/Jun/19
$${x}=\mathrm{20}°,\:{see}\:{also}\:{Q}\mathrm{43728}. \\ $$
Answered by peter frank last updated on 01/Jun/19
$${x}=\mathrm{40}° \\ $$
Answered by mr W last updated on 02/Jun/19
Commented by mr W last updated on 02/Jun/19
$${draw}\:{DF}//{AB} \\ $$$$\Delta{DFG}\:{is}\:{equilateral},\:\Rightarrow{DF}={GF} \\ $$$$\Delta{AFC}\:{is}\:{isosceles},\:\Rightarrow{AF}={CF} \\ $$$$\Delta{AGC}\equiv\Delta{CEA},\:\Rightarrow{AG}={CE} \\ $$$${AF}−{AG}={CF}−{CE} \\ $$$$\Rightarrow{GF}={EF} \\ $$$$\Rightarrow\Delta{DFE}\:{is}\:{isosceles}, \\ $$$$\Rightarrow\angle{DEF}=\angle{EDF}=\frac{\mathrm{180}−\angle{DEF}}{\mathrm{2}}=\frac{\mathrm{180}−\mathrm{80}}{\mathrm{2}}=\mathrm{50}° \\ $$$$\angle{AEB}=\mathrm{180}−\mathrm{70}−\mathrm{80}=\mathrm{30}° \\ $$$${x}=\angle{DEF}−\angle{AEB}=\mathrm{50}−\mathrm{30}=\mathrm{20}° \\ $$
Commented by otchereabdullai@gmail.com last updated on 11/Apr/20
$$\mathrm{wow} \\ $$