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Junior high school mathematics-circle (perpendicular to the diameter of the string)
(2) Solution: Connect BD. ABD is a right triangle with AD= 10 and AB=8, so BD=6. △ABD area = AB * BD/2 = AD * BE/2。 So be = AB * BD/AD = 8× 6/ 10 = 4.8, BC=2BE=2×4.8=9.6.

2. As shown in the figure, P is a point on the chord AB, CP⊥OP crosses ⊙O at point C, AB=8, AP: Pb = 1: 3, and find the length of PC.

Solution: Shall we draw three lines first? Od? OA? 、OC? Jean Od⊥AB

Get the right triangle OAd,? Right triangle OPd, right triangle OPC

AB = 8,AP:PB= 1:3

∴AP:(AP+PB)= 1:( 1+3)? [ratio theorem]

that is

AP:AB= 1:4

AP=AB/4=8/4=2。

BP=AB-AP=8-2=6,

∵Od⊥AB

∴Ad=Bd (vertical diameter chord splitting theorem)

It can be concluded that d is the midpoint of straight line AB.

AB=8 then AD=AB/2=4.

Then, according to the Pythagorean theorem of the right triangle edge

Right triangle oad: OA 2 = OD 2+4 2

? oa^2=od^2+ 16

Right triangle OPD: OP 2 = OD 2+2 * 2

? od^2=op^2-4

Right triangle OPC: oc 2 = op 2+PC 2

? pc^2=oc^2-op^2

Circle o radius OA=OC:? pc^2=od^2+ 16-op^2

? pc^2=(op^2-4)+ 16-op^2

? pc^2= 12

? pc=√? 12=2√3≈2× 1.732

? pc≈3.46