Question Number 186376 by Rupesh123 last updated on 04/Feb/23
Answered by HeferH last updated on 04/Feb/23
Commented by HeferH last updated on 04/Feb/23
$$\mathrm{1}.{move}\:{triangle}\:{ADB}\: \\ $$
Commented by HeferH last updated on 04/Feb/23
Commented by HeferH last updated on 04/Feb/23
$$\mathrm{2}.\:{now}\:{ADH}\:{is}\:{isosceles},\:{and}\:{also}\:{AHC}\:{is} \\ $$$${isosceles}\:\left(\mathrm{10}°,\:\mathrm{10}°,\:\mathrm{160}°\right)\: \\ $$$$\: \\ $$
Commented by HeferH last updated on 04/Feb/23
Commented by HeferH last updated on 04/Feb/23
$$\mathrm{3}.\:\bigtriangleup\:{ABH}\:\cong\:\bigtriangleup\:{DCB} \\ $$
Commented by HeferH last updated on 04/Feb/23
Commented by HeferH last updated on 04/Feb/23
$$\mathrm{4}.\:{now}\:{we}\:{want}\:{to}\:{find}\:\angle{ABH}\:=\:\angle\:{BCD}\:=\:{x}\: \\ $$
Commented by HeferH last updated on 04/Feb/23
Commented by HeferH last updated on 04/Feb/23
$$\mathrm{5}.\:{Since}\:\angle{BAD}\:=\:\mathrm{2}\angle{ABD}\:{we}\:{can}\:{draw}\: \\ $$$$\:{this}\:{segment} \\ $$
Commented by HeferH last updated on 04/Feb/23
Commented by HeferH last updated on 04/Feb/23
$$\mathrm{6}.\:{we}\:{have}\:{an}\:{equilateral}\:{triangle}\:{that}\: \\ $$$$\:{leads}\:{to}\:{another}\:{isosceles}\:{triangle} \\ $$$$\:\left(\mathrm{100}°,\:\mathrm{40}°,\mathrm{40}°\right)\: \\ $$
Commented by HeferH last updated on 04/Feb/23
Commented by HeferH last updated on 04/Feb/23
$$\mathrm{7}.\:{finally}\:\angle{ABH}\:=\:\mathrm{40}°\:=\:{x}\: \\ $$
Commented by HeferH last updated on 04/Feb/23
$$:\mid \\ $$