от Добромир Глухаров » 18 Фев 2020, 20:23
Нека $\angle CAB=2\alpha\Rightarrow\angle CAP=\alpha,\angle CBA=90^\circ-2\alpha$
$\angle ACD=90^\circ-2\alpha,\angle ACP=\angle ACD+\angle DCP=90^\circ-2\alpha+\frac{1}{2}\left(90^\circ-\angle ABC\right)=90^\circ-2\alpha+\frac{1}{2}\left(90^\circ-\left(90^\circ-2\alpha\right)\right)=90^\circ-2\alpha+\frac{1}{2}\left(2\alpha\right)=90^\circ-\alpha$
$\angle APC=180^\circ-\angle CAP-\angle ACP=180^\circ-\alpha-\left(90^\circ-\alpha\right)=90^\circ$