精题例解·合作探究
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1-1 ∠A ∠AED
AC/AE BC/DE
2-1 A
3-1 证明: ∵ AB ∥ DN,
∴ △AMB ∽ △NMD,
∴ AM/MN = BM/DM.
又 AD ∥ BP,∴ △BMP ∽ △DMA,
∴ MP/AM = BM/DM
∴ AM/MN = MP/AM
∴ AM² = MN·MP.
4-1 解:相似,理由;
∴△ABC ∽ △B′A′C′
5-1 解:由已知得∠C = ∠D = 90°,AD = BC = 1,PD = PC = 0.5.
Rt△ADP ∽ Rt△QCP,只需AD/QC = PD/PC成立,
即1/CQ = 0.5/0.5,解得CQ = 1,
即Q与B重合,BQ = 0.
若Rt△ADP ∽ Rt△PCQ,只需AD/QC = PD/PC成立,
即 Q 与 B 重合,解得CQ = 0.25,
则BQ = 1 - 0.25 = 0. 75.
综上可得,当BQ = O或 BQ = 0.75时,△ADP与△QCP相似.
6-1 证明:如图D-27-1所示,连接BE
∵ AE是⊙O的直径,
∴ ∠ABE = 90°.
又 AD ⊥ BC,
∴ ∠ADC = 90°.
∴ ∠ABE = ∠ADC..
在△ABF_和△ADC中,∠ABE = ∠ADC,
∠E = ∠C(同弧所对的圆周角相等).
∴ △ABE ∽ △ADC
∴ AB/AD = AE/AC
∴ AB·AC = AD·AE