|
|
乌贼
Posted 2023-11-19 23:56
Last edited by 乌贼 2023-11-22 12:42如图：
延长$ BA $交圆$ A $于$ D $。\[ \angle IPB=\angle IQB=\angle IBQ\riff\angle IRB=\angle IBP\riff\triangle IPB\sim \triangle IBR\riff IC^2=IQ^2=IP\cdot IR\riff\triangle ICP\sim \triangle IRC\riff\angle IRC=\angle ICP \]所以\[ \angle BRC=\angle BRI+\angle CRI=\angle IBP+\angle ICP=\angle IBP+\angle ICB+\angle BCP=\angle IBP+\angle IBD+\angle BDP=90\du \]
|
|
