1. Analysis: 1) By connecting AD, that is, the height at the base of the isosceles triangle ABC, we can get ∠CAD=∠BAD according to the characteristics of the three lines of the isosceles triangle, and ∠DEB=∠DBE according to the theorem of the circle angle, and we can prove that DE = DB.

② In this problem, BEcause BE⊥AC, then be is the height ABC of the AC side in the triangle, and AC can be obtained by different area representations. BE=CB？ Advertising. Then find the length of BE.

2. Solution: Solution: (1)DE=BD

Evidence: link advertisement, then AD⊥BC.

In the isosceles triangle ABC, AD⊥BC

∴∠ CAD =∠∠ bad (isosceles triangle connected by three lines)

∠∠CAD =∠DBE，∠BAD=∠DEB

∴∠DEB=∠DBE

∴de=bd；

(2)∫AB = 5，BD= BC=3

∴AD=4 solution: (1)DE=BD

Evidence: link advertisement, then AD⊥BC.

In the isosceles triangle ABC, AD⊥BC

∴∠ CAD =∠∠ bad (isosceles triangle connected by three lines)

∠∠CAD =∠DBE，∠BAD=∠DEB

∴∠DEB=∠DBE

∴de=bd；

(2)∫AB = 5，BD= BC=3

∴AD=4

AB = AC = 5

∴AC？ BE=CB？ advertisement

∴BE=4.8.

AB = AC = 5

∴AC？ BE=CB？ advertisement

∴BE=4.8.

the second question

Proof: (1) Connect OT;

Pq cut ⊙ o to t,

∴OT⊥PQ，

And ∵AC⊥PQ,

∴OT‖AC，

∴∠tac=∠ato；

OT = OA，

∴∠ATO=∠OAT，

∴∠OAT=∠TAC，

That is to say, in the stock ∠ BAC

(2) point o is OM⊥AC in m,

∴am=md= = 1；

∠ OTC =∠ Act =∠ OMC = 90，

∴ Quadrilateral OTCM is rectangular,

∴OM=TC=，

Ⅶ in Rt△AOM

;

That is, the radius of ⊙O is 2.