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
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/03 00:47:15
![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](/uploads/image/z/14908303-55-3.jpg?t=P%E6%98%AF%E5%9B%9B%E8%BE%B9%E5%BD%A2ABCD%E5%86%85%E4%B8%80%E7%82%B9%2C%E4%B8%94PA%3APB%3APC%3D2%3A1%3A3%E8%AF%81%E6%98%8E%E8%A7%92APB%E4%B8%BA135%C2%B0.JJJJJJJJJJJJJJJJJJJJJJ%E6%AD%BB%E4%BA%86)
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°