Question Number 223666 by MASANJAJJ last updated on 01/Aug/25
Answered by mr W last updated on 01/Aug/25
Commented by mr W last updated on 02/Aug/25
$${i}\:{can}'{t}\:{remember}\:{in}\:{which}\:{year},\:{but} \\ $$$${it}\:{was}\:{a}\:{long}\:{time}\:{ago}.\: \\ $$
Commented by mr W last updated on 02/Aug/25
$${AB}//{DC}\: \\ $$$$\Rightarrow\angle{ABC}=\angle{BAD}=\mathrm{180}°−\mathrm{30}°−\mathrm{50}°=\mathrm{100}° \\ $$
Commented by fantastic last updated on 02/Aug/25
$${Sir}\:{i}\:{have}\:{a}\:{question}. \\ $$$${Just}\:{to}\:{ask} \\ $$$${which}\:{year}\:{did}\:{you}\:{join}\:{this}\:{app}?? \\ $$$$\left.{please}\:{tell}\:{me}\::\right) \\ $$
Commented by necx122 last updated on 02/Aug/25
I've been here since 2015 or 2016 and Mr. W has been the most active member alongside Ajfour. These are exceptional men here
Commented by fantastic last updated on 02/Aug/25
$${Mr}.{W}\:\:−\:{the}\:{veteran} \\ $$
Answered by fantastic last updated on 02/Aug/25
Commented by fantastic last updated on 02/Aug/25
$$\measuredangle{FCE}=\mathrm{180}^{\mathrm{0}} −\mathrm{50}^{\mathrm{0}} −\mathrm{30}^{\mathrm{0}} =\mathrm{100}^{\mathrm{0}} \\ $$$$\therefore\measuredangle{FHE}=\mathrm{180}^{\mathrm{0}} −\mathrm{100}^{\mathrm{0}} =\mathrm{80}^{\mathrm{0}} \\ $$$$\measuredangle{FEC}=\measuredangle{FCG}=\mathrm{30}^{\mathrm{0}} \\ $$$$\because{CE}\parallel{FH}\:{so}\:\measuredangle{FEC}=\measuredangle{EFH}=\mathrm{30}^{\mathrm{0}} \\ $$$${So}\:\measuredangle{HEF}=\mathrm{180}^{\mathrm{0}} −\mathrm{80}^{\mathrm{0}} −\mathrm{30}^{\mathrm{0}} =\mathrm{70}^{\mathrm{0}} \\ $$$$\therefore\measuredangle{HEC}=\measuredangle{HEF}+\measuredangle{FEC}=\mathrm{70}^{\mathrm{0}} +\mathrm{30}^{\mathrm{0}} =\mathrm{100}^{\mathrm{0}} \checkmark \\ $$