∫OE∨AB，

∴∠COE=∠CAD,∠EOD=∠ODA，

OA = OD，

∴∠OAD=∠ODA，

∴∠COE=∠DOE，

In △COE and △DOE,

CO=OD∠COE=∠OE=OEDOE，

∴△COE≌△DOE(SAS)，

∴∠ODE=∠OCE=90，

∴ED⊥OD，

∴ED is the tangent of circle o;

(2) Connect the CD and pay the OE to M,

In Rt△ODE,

OD = 32，DE=2，

∴OE= OD2+DE2= ( 32)2+22= 52，

∫DE∨AB，

∴△COE∽△CAB，

∴ Oka = OEAB,

∴AB=5，

∫AC is the diameter,

∴∠ADC=90，

∴cos∠BAC= ACAB= ADAC= 35，

∴AD= 95，

∴CD= AC2-AD2= 125，

∫EF∨AB，

∴ CMDM= OCOA，

∴CM=DM= 12CD= 65，

∴ef=oe+of=4,bd=ab-ad=5- 95 = 165，

∴S△ADF=S trapezoidal ABEF-S trapezoidal DBEF= 12(AB+EF)? 6? 1DM- 12(BD+EF)？ 6? 1DM = 12×(5+4)×65- 12×( 165+4)×65 = 2725。

The area of ∴△ADF is 2725 square kilometers. Comments: This topic examines the judgment and nature of the tangent of a circle, the nature of a right triangle, the judgment and nature of similar triangles, and the judgment and nature of congruent triangles. This question is comprehensive and difficult. The key to solving the problem is to pay attention to the application of the idea of combining numbers with shapes. Pay attention to the practice of auxiliary lines. Answer: zcx teacher hides and analyzes online training collection questions, downloads questions, exam baskets, and more papers. Recommended Test Paper 20 10 * * * 5 Comments on Mathematical Simulation Test Paper of Senior High School Entrance Examination in Wuhan City, Hubei Province, with an average score of 10 (full score 10). Click to display comments/corrections.

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