矩形ABCD,CD与以AB为直径的半圆相切于F,O为圆心,AC交半圆于P,过P作PE垂直AB于E,AB=5,求PE的长

Question

矩形ABCD,CD与以AB为直径的半圆相切于F,O为圆心,AC交半圆于P,过P作PE垂直AB于E,AB=5,求PE的长

Answer 1

连接OF,CD与半圆相切于F,∴OF⊥CD,
OF//BC//AD,
BC=AD=OF=AB/2=5/2;
PE⊥BC,
连接BP,AB为直径,∴角APB=90°,角EAP=∠BAC=∠EPB,
RT△EAP∽RT△BAC∽RT△EPB,[AA]
PE:AE=BC:AB,AE=AB*PE/BC=5*PE/(5/2)=2PE;
EB:PE=BC:AB,EB=BC*PE/AB=(5/2)PE/5=PE/2;
AE+EB=AB
2PE+PE/2=5
PE=5/(5/2)=2;