Question

如图,已知矩形ABCD中,AB＝4/3BC,O是矩形ABCD的中心,、（对角线的交点）,过点O作OE⊥BC于F,得矩形B

Answer 1

1.AE = 4/3CF,垂直

2.AB/BC = BE/BF=4/3 ,角ABE= 90度-角EBC= 角CBF,所以三角形ABE相似于三角形CBF,

AE/CF =4/3,角EAB= 角FCB,设AE与CB交于G,与CF交于H,角EAB+角AGB=90度 = 角FCB+角CGH 所以,角CHG= 90度,所以AE与CF垂直

3.角FCB= 角EAB,AEO成一条直线,BF/BC = 1/2; BP/AB = 1/2,AD =4,AB = 16/3,BP = 8/3;

CP = 4 -8/3 = 4/3;