如图,在△ABC中,∠ACB=90°,AC=BC,P为△ABC内一点,且PA=3,PB=1,CD=PC=2,CD⊥CP.求∠BPC图 http://dl.zhishi.sina.com.cn/upload/79/85/51/1582798551.1234770495.jpg

如图,在△ABC中,∠ACB=90°,AC=BC,P为△ABC内一点,且PA=3,PB=1,CD=PC=2,CD⊥CP.求∠BPC图 http://dl.zhishi.sina.com.cn/upload/79/85/51/1582798551.1234770495.jpg
如图,在△ABC中,∠ACB=90°,AC=BC,P为△ABC内一点,且PA=3,PB=1,CD=PC=2,CD⊥CP.求∠BPC
图 http://dl.zhishi.sina.com.cn/upload/79/85/51/1582798551.1234770495.jpg

如图,在△ABC中,∠ACB=90°,AC=BC,P为△ABC内一点,且PA=3,PB=1,CD=PC=2,CD⊥CP.求∠BPC图 http://dl.zhishi.sina.com.cn/upload/79/85/51/1582798551.1234770495.jpg
∵AC⊥BC,PC⊥CD
∴∠ACB=∠PCD=90°
∵∠ACP=∠ACB-∠BCP=∠PCD-∠BCP.
AC=BC,CP=CD
∴△ACP≌△BCD
AP=BD=3
PD=√(CP^2+CD^2)=2√2
BD^2=9=1^2+(2√2)^2=CP^2+CD^2 或 cosBPD=(BP^2+DP^2-BD^2)/(2BP*DP)=0
∴∠BPD=90°
∵PC⊥CD,CP=CD
∴∠CPD=45°
∴∠BPC=∠CPD+∠BPD=135°