Question

初三数学，求解。

Answer 1

1)
|OE|^2 = |OD|^2 - |DE|^2 (勾股定理)
|OD| = |AB|/2 = 6
|OE|^2 = 36 - 32 = 4
|OE| = 2

2)
延长 DP 交圆与另一点 C'
角DPA = 角BPC' (对顶角相等)
显然 |CP| = |C'P|
而 OE 垂直于弦 C'D, 因此它是 C'D 的垂直平分线,
|DE| = |C'E| = |C'P| + |PE| = |CP| + |PE|
因此, PD + PC = DE + EP + PC = 2 DE

3)
|DP| = 2 DE - PC = 8sqrt(2) - 3sqrt(2) = 5sqrt(2)
sin角CPB = sin角OPE = |OE|/|OP|
|OE| = 2, |PE| = sqrt(2)
|OP| = sqrt(6)
sin角CPB = sqrt(6)/3