摘要:23.(1)△OBD∽△PAD. 证明 ∵ PA.PB是⊙O的切线.∴ OA⊥PA.OB⊥PB.∴ ∠OAP =∠OBD = 90°. 又∠D =∠D.∴ △OBD∽△PAD. (2) ∵ ∠P = 45°. ∴ ∠DOB = 45°.∴ △OBD.△PAD均是等腰直角三角形. 从而 PD =PA.BD = OB. 又 ∵ PA = 2 +.PA = PB. ∴ BD = OB = PD-PB =PA-PA=(-1)PA=(-1)(2+)=. 故 S阴影 = S△OBD-S扇形 ===.

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