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How to do 20 math problems in 20 10 Hengyang senior high school entrance examination?
(1) Connect BD, and get right triangles ABD and BCD according to the right angle of the circle opposite to the diameter. According to the judgment theorem of tangent, we know that BC is the tangent of a circle, and we can get BE=DE by combining the tangent length theorem, and then we can prove DE = CE according to the congruence angles of equilateral corners and equilateral corners.

(2) In the right triangle ABC, calculate its three sides according to the concept of acute trigonometric function and Pythagorean theorem, and then calculate according to similar triangles's judgment and properties.

Solution: (1) Proof: Connect BD,

∫AB is the diameter, ∠ ABC = 90,

∴BC is the tangent of⊙ O, and∠ ∠BDC = 90°.

And DE is the tangent of ⊙O,

∴DE=BE.

∴∠EBD=∠EDB.

And < DCE+< EBD = < CDE+< EDB = 90,

∴∠DCE=∠CDE,

∴DE=CE.

So DE= BC ..

(2) From (1), BC = 2DE = 4.

In Rt△ABC, AB=BCtanC=4× radical number 5/2 =2 radical number 5,

So AC= 6 (Pythagorean theorem).

∠∠ADB =∠ABC = 90,∠A=∠A,

∴ABD∽△ACB.

∴AD/AB=AB/AC

∴AD/ root number 5= root number 5/6

The solution is AD= 10/3.