Question Number 179653 by cherokeesay last updated on 31/Oct/22
Answered by mr W last updated on 31/Oct/22
Commented by mr W last updated on 31/Oct/22
$$\angle{ABD}=\angle{ACD}=\theta \\ $$$$\angle{CBH}=\varphi=\mathrm{90}°−\theta \\ $$$$\angle{AGP}=\mathrm{180}°−\mathrm{2}\phi \\ $$$$\angle{AGP}=\mathrm{2}×\angle{ABD}=\mathrm{2}\theta \\ $$$$\mathrm{180}°−\mathrm{2}\phi=\mathrm{2}\theta \\ $$$$\Rightarrow\phi=\mathrm{90}°−\theta=\varphi \\ $$$${PG}\:{and}\:{PH}\:{are}\:{in}\:{the}\:{same}\:{line}, \\ $$$${i}.{e}.\:{G},{P},{H}\:{are}\:{collinear}. \\ $$