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17. 解:法军能命中. 理由如下:
根据题意ꎬ得 AB = POꎬ∠A =∠P.
又∵ AB⊥BOꎬPO⊥BQꎬ∴ ∠ABO =∠POQ = 90°.
在△ABO 和△POQ 中ꎬ∵ ∠A =∠PꎬAB = POꎬ∠ABO =∠POQ = 90°.
∴ △ABO≌△POQ(ASA) . ∴ BO = OQ.
因此ꎬ按照 BO 的距离炮轰德军时ꎬ炮弹恰好落入德军 Q 处. 故法军能命中目标.
18. 解:(1)点 A10的坐标为(1ꎬ - 5) .
(2)观察发现:A1(2ꎬ0)ꎬA5(4ꎬ0)ꎬA9(6ꎬ0)ꎬꎬ
A3(0ꎬ0)ꎬA7( - 2ꎬ0)ꎬA11( - 4ꎬ0)ꎬ
∴ A4n + 1(2n + 2ꎬ0)ꎬA4n - 1( - 2n + 2ꎬ0)(n 为自然数) .
∵ 2 021 = 505 × 4 + 1ꎬ∴ A2 021的坐标为(1 012ꎬ0) .
19. 解:(1)∵ ∠A ∶ ∠ABC = 3 ∶ 4ꎬ∴ 设∠A = 3kꎬ∠ABC = 4k.
又∵ ∠ACD =∠A +∠ABC = 140°ꎬ
∴ 3k + 4k = 140°. 解得 k = 20°. ∴ ∠A = 3k = 60°.
(2)∵ ∠MCD 是△MBC 的外角ꎬ∴ ∠M =∠MCD -∠MBC.
同理可得ꎬ∠A =∠ACD -∠ABC.
∵ CMꎬBM 分别平分∠ACDꎬ∠ABCꎬ∴ ∠MCD = 12 ∠ACDꎬ∠MBC =
1
2 ∠ABC.
∴ ∠M = 12 (∠ACD -∠ABC) =
1
2 ∠A.
∵ CP⊥BMꎬ∴ ∠PCM = 90° -∠M = 90° - 12 ∠A.
20. 解:(1)设 y2 关于 x 的函数表达式是 y2 = kx + b.
根据题意ꎬ得 20k + b = 0ꎬ40k + b = 4.{ 解得
k = 0. 2ꎬ
b = - 4.{
∴ y2 关于 x 的函数表达式是 y2 = 0. 2x - 4.
(2)由图象可知ꎬ步行的学生的速度为 4 ÷ 40 = 0. 1(千米 /分钟) .
∴ 步行同学到达百花公园的时间为 6 ÷ 0. 1 = 60(分钟) .
当 y2 = 6 时ꎬ6 = 0. 2x - 4. 解得 x = 50. ∵ 50 < 60ꎬ∴ 60 - 50 = 10(分钟) .
答:骑自行车的学生先到达百花公园ꎬ先到了 10 分钟.
21. (1)证明:∵ AIꎬBI 分别平分∠BACꎬ∠ABCꎬ
∴ ∠BAI = 12 ∠BACꎬ∠ABI =
1
2 ∠ABC.
∴ ∠BAI +∠ABI = 12 (∠BAC +∠ABC) =
1
2 (180° -∠ACB) = 90° -
1
2 ∠ACB.
∴ 在△ABI 中ꎬ∠AIB = 180° - (∠BAI + ∠ABI) = 180° - (90° - 12 ∠ACB) = 90° +
1
2 ∠ACB.
∵ CI 平分∠ACBꎬ∴ ∠DCI = 12 ∠ACB. ∵ DI⊥ICꎬ∴ ∠DIC = 90°.
∴ ∠ADI =∠DIC +∠DCI = 90° + 12 ∠ACB. ∴ ∠AIB =∠ADI.
(2)解:①DI∥CF. 理由如下:
∵ ∠IDC = 90° -∠DCI = 90° - 12 ∠ACBꎬCF 平分∠ACEꎬ
∴ ∠ACF = 12 ∠ACE =
1
2 (180° -∠ACB) = 90° -
1
2 ∠ACB.
∴ ∠IDC =∠ACF. ∴ DI∥CF.
②∵ ∠ACE =∠ABC +∠BACꎬ∴ ∠ACE -∠ABC =∠BAC = 70°.
∵ ∠FCE =∠FBC +∠Fꎬ∴ ∠F =∠FCE -∠FBC.
∵ ∠FCE = 12 ∠ACEꎬ∠FBC =
1
2 ∠ABCꎬ
∴ ∠F = 12 ∠ACE -
1
2 ∠ABC =
1
2 (∠ACE -∠ABC) = 35°.
22. (1)证明:∵ ∠BAC =∠BAD +∠DACꎬ∠DAE =∠DAC +∠CAEꎬ∠BAC =∠DAEꎬ
∴ ∠BAD =∠CAE.
在△ABD 和△ACE 中ꎬ
AB = ACꎬ
∠BAD =∠CAEꎬ
AD = AEꎬ{ ∴ △ABD≌△ACE(SAS) . ∴ CE = BD.
(2)解:α + β = 180°. 证明如下:由(1)知ꎬ△ABD≌△ACE. ∴ ∠ACE =∠Bꎬ
又∠B +∠ACB + α = 180°ꎬ∴ ∠ACE +∠ACB + α = 180°.
又∵ ∠ACE +∠ACB =∠DCE = βꎬ∴ α + β = 180°.
(3)解:α 与 β 之间的数量关系是 α = β.
23. 解:(1)设直线 AB 的函数表达