Question Number 187828 by Rupesh123 last updated on 22/Feb/23
Answered by HeferH last updated on 22/Feb/23
Commented by HeferH last updated on 22/Feb/23
$${i}.\:\mathrm{3}\alpha\:+\:\beta\:+\omega\:=\:\mathrm{180}° \\ $$$${ii}.\:\:\omega\:+\:\alpha\:=\:\mathrm{60}° \\ $$$${iii}.\:\alpha=\:\mathrm{60}°\:−\:\omega \\ $$$$\:\mathrm{3}\left(\mathrm{60}°−\omega\right)\:+\:\beta\:+\omega\:=\:\mathrm{180}° \\ $$$$\:\mathrm{180}°\:−\mathrm{3}\omega\:+\:\beta\:+\omega\:=\:\mathrm{180}° \\ $$$$\:\beta\:=\:\mathrm{2}\omega \\ $$$$\:\angle{ACB}\:=\:\mathrm{3}\omega \\ $$$$\:\angle{ACE}\:=\:\omega\: \\ $$$$\:\angle{ACB}\:=\:\mathrm{3}\centerdot\angle{ACE}\:\checkmark \\ $$
Commented by Rupesh123 last updated on 23/Feb/23
Very nice approach, sir!