A(10, 0), B(10, 8), C(0, 8), D(6, 0), E(10, 3)
CE解析式: y = 8 - x/2
P(16, 0)
直线l: y = k(x - 6)
(i)
凭直觉可以看出直线l与CE垂直时是一个解.此时二者均为直角三角形,只需∠DPQ = ∠CRQ即可
tan∠DPQ = OC/OP = 8/16 = 1/2
此时k = -1/(-1/2) = 2
与y轴交点为R(0, -12)
与CE交点为Q(8, 4)
CQ = 4√5, QR = 8√5
tan∠CRQ = 4√5/(8√5) = 1/2
∠DPQ = ∠CRQ
二者相似, y = 2x - 12
(ii)
这时将三角形ODR沿着x轴向上翻折,得三角形ODR', DR'与CE交点为Q'
显然∠CR'Q' = ∠DPQ'; ∠DQ'P = ∠CQ'R' (对顶角)
此时二者也相似.
R'(0, 12)
k' = (12 - 0)/(0 - 6) = -2
y = -2x + 12