P是四边形ABCD内一点,且PA:PB:PC=2:1:3证明角APB为135°.JJJJJJJJJJJJJJJJJJJJJJ死了

P是四边形ABCD内一点,且PA:PB:PC=2:1:3证明角APB为135°.JJJJJJJJJJJJJJJJJJJJJJ死了
P是四边形ABCD内一点,且PA:PB:PC=2:1:3证明角APB为135°
.JJJJJJJJJJJJJJJJJJJJJJ死了

P是四边形ABCD内一点,且PA:PB:PC=2:1:3证明角APB为135°.JJJJJJJJJJJJJJJJJJJJJJ死了
题目是不是错了,应该是正方形ABCD吧?
是的话,这样做：
将△BAP绕B点旋转90°使BA与BC重合,P点旋转后到Q点,连接PQ
因为△BAP≌△BCQ
所以AP＝CQ,BP＝BQ,∠ABP＝∠CBQ,∠BPA＝∠BQC
因为四边形DCBA是正方形
所以∠CBA＝90°
所以∠ABP＋∠CBP＝90°
所以∠CBQ＋∠CBP＝90°
即∠PBQ＝90°
所以△BPQ是等腰直角三角形
所以PQ＝√2*BP,∠BQP＝45°
因为PA=1,PB=2,PC=3
所以PQ＝2√2,CQ＝1
所以CP^2＝9,PQ^2＋CQ^2＝8＋K＝9
所以CP^2＝PQ^2＋CQ^2
所以△CPQ是直角三角形且∠CQA＝90°
所以∠BQC＝90°＋45°＝135°
所以∠BPA＝∠BQC＝135°