如图,在平面直角坐标系中,点A,B的坐标分别为(-5,0)和(5,0),以AB为直径在x轴的上方作半圆O,点C是该半圆上第一象限内的一个动点,连结AC,BC,并延长BC至点D,使BC=CD,过点D作x轴的垂线,分别交x轴,线段A

Question

如图,在平面直角坐标系中,点A,B的坐标分别为(-5,0)和(5,0),
以AB为直径在x轴的上方作半圆O,点C是该半圆上第一象限内的一个动点,连结AC,BC,并延长BC至点D,使BC=CD,过点D作x轴的垂线,分别交x轴,线段AC于点E,F,点E为垂足,连结OF.(1)当∠CAB=30°时,求的长; (2)当AE=6时,求弦BC的长; (3)在点C运动的过程中,是否存在以点O,E,F为顶点的三角形与△DEB相似,若存在,请求出此时点E的坐标;若不存在,请说明理由.

Answer 1

(1)题不全
(2)
从C向x轴做垂线,垂足C'
AE = 6, BE = AB - AE = 10 - 6 = 4
CC'平分BD和BE,BC' = BE/2 = 2, C'(3, 0)
圆x² + y² = 25
代入x = 3, y = 4 (y >0, C与C'横坐标相同)
C(3, 4)
BC= √[(3 - 5)² + (4 - 0)²] = 2√5
(3)
C(p, q), p² + q² = 25, 0 < p < 5, 0 < q < 5
C'(p, 0), C'B = 5 - p, EB = 2C'B = 10 - 2p
E(2p - 5, 0)
D(2p - 5, 2q)
AC: (y - 0)/(q - 0) = (x + 5)/(p + 5)
x = 2p - 5, y = 2pq/(p + 5)
F(2p - 5, 2pq/(p + 5))
tan∠EOF = EF/OE = 2pq/[(p + 5)|(2p - 5)|] (因为2p - 5可以为负值)
tan∠BDE = EB/DE = (10 - 2p)/(2q) = (5 - p)/q
tan∠DBE = DE/BE = 2q/(10 - 2p) = q/(5 - p)
(i)如∠EOF和∠BDE为对应角2pq/[(p + 5)|(2p - 5)|] = (5 - p)/q
2p/|2p - 5| = (25 - p²)/q² = q²/q² = 1
2p = |2p - 5|
p = 5/4, E(-5/2, 0)
(ii) 如∠EOF和∠DBE为对应角
2pq/[(p + 5)|(2p - 5)|] = q/(5 - p)
(a) p >= 5/2
4p² - 5p - 25 = 0
p = 5(1 + √17)/8
E(5(√17 - 3)/4, 0)
(b) 0 < p < 5/2
p = 5/3
E(-5/3, 0)