Prove that BC = 6△ABC in (1)Rt is known? AB= 10, Pythagorean theorem: ab? =AC？ +BC means 10=AC? +6 ? Solution? AC=8

So what? cosA=8/ 10=4/5

Connect the CD because BC is the diameter.

So △BDC is Rt△, which is CD⊥AB.

So Rt△BDC∽Rt△ABC

Have BD/BC=BC/AB? Known BC=6? AB= 10 becomes BD/6=6/ 10?

BD=3.6

So AD=AB-BD= 10-3.6=6.4.

Therefore, AM=AC-CM=8-4=4, AD=6.4, COSA = 4/5 in △ ADM.

Is there a DM with cosine theorem? =AD？ +AM？ -2*AD*AM*cosA=6.4？ +4? -2*6.4*4*4/5= 16

So DM=√DM? =√ 16=4(-4 rounded)

So DM=CM=4

Therefore, in the connection OM, OC=OD= radius = 3 OM, △OCM and △ODM are * * *, and DM=CM.

So △ OCM △ ODM (? Side by side) that is, ∠OCM=∠ODM and ∠ OCM = ∠ ACB = 90.

So ∠ ODM = 90

DM⊥OD

So the straight line DM is the tangent of ⊙O;

(2) as shown in figure m? O is the midpoint of the AC side and BC side of Rt△ABC respectively.

So OM∨AB∠A =? ∠OMC=？ ∠OMD

That is ∠ cmd = 2 ∠ a.

Understand OCMD four-point circle easily.

Therefore, EOC = cmd = 2a.

cos∠EOC=cos2A

Certification cosA=4/5

Cos2A=2cos？ A- 1

cos∠EOC=2cos？ A- 1=2*(4/5)？ - 1=0.28

△EOC cos∠EOC=0.28

OC=3

cos∠EOC=3/OE=0.28。

CE/OE=sin∠EOC=√( 1-cos？ ∠EOC)=√( 1-(0.28)？ )=√0.92 16=0.96