Solving problems of geometric circle in mathematics senior high school entrance examination

Solution: (1) Take the midpoint o of BD, because DE⊥BE.

So o is the center of the circumscribed circle of △DBE, and DB is the diameter.

Connect OE, so: OE=OB=OD.

So: ∠OEB=∠OBE, ∠ODE=∠OED.

Because it is divided equally ∠ABC

Therefore: ∠OEB =∠ OBE =∠EBC.

Because ∠ c = 90.

Therefore: ∠ CEB+∠ EBC = 90.

Therefore: ∠ OEC = ∠ OEB+∠ CEB = ∠ CEB+∠ EBC = 90.

Therefore, AC is the tangent of the circumscribed circle of △DBE.

(2) Because AC is the tangent of the circumscribed circle of △DBE, ∠ C = 90.

Therefore, AEO = 90 = AED+OED = AED+ODE = OBE+ODE, OE ∠ BC.

Therefore: ∠AED =∠ obei.

Because ∠A=∠A

So: △AED∽△ Abe

So: AE /AB=AD/AE=DE/BE.

Because AD=6, AE=6√2.

Therefore: AB= 12, DE/BE =√2/2.

Therefore: BD=6, OE=3, AO=9.

Because BC

Therefore: OE/BC=AO/AB.

Therefore: BC=4.