PE⊥PA
∠APE=90
∠APB+∠CPE=90
又因为正方形ABCD
∠ABP=∠ECP=90
∠APB+∠PAB=90
∠CPE=∠PAB
△ABP∽△PEC
又因为BP=5,BC=AB=12
PC=7
CP/AB=EC/BP
EC=35/12
DE=CD-EC=109/12
已知:正方形ABCD的边长为12,点P在BC上,BP=5,PE垂直于AP,交CD于E,点DE的长为多少?
已知:正方形ABCD的边长为12,点P在BC上,BP=5,PE垂直于AP,交CD于E,点DE的长为多少?
数学人气:174 ℃时间:2019-11-21 18:37:01
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