CDM concept photography

Solution: If the diameter AB is perpendicular to the chord CF, then: arc AF= arc AC; And OC=OB*OA.

That is, 4=OB*2, OB=8, AB= 10, and the radius of the circle is AB/2=5.

Connect BN, O'N, CN, then ∠ nbm = ∠ CDM; ∠NMB=∠CMD。

∫ arc AC= arc CD; CM = communication.

∴AC=CD=CM,∠CMD=∠CDM (equilateral and equilateral).

∴∠NMB=∠NBM (equivalent substitution), MN=BN.

Let DP∑CB intersect with O' in P, then arc PB= arc CD= arc AC= arc AF; PDM=∠CMD=∠CDM。

∴ arc PBN= arc tank; Arc PB= arc AC.

So the arc BN= the arc AN, that is, the point N is the midpoint of the semicircle BNA. If there is no' connectivity, then NO'⊥BO'.

∴mn=bn=√(o'b+o'n)=5√2；

If AB is the diameter, then ∠ ACB = 90, BC = √ (AB? -AC? )=√( 100-20)=4√5;

And CM=AC=2√5, then BM=BC-CM=2√5.

According to the symphony theorem (or ⊿BMN∽⊿DMC), DM*MN=BM*CM can be obtained.

That is DM*(5√2)=(2√5)*(2√5),? DM = 2ì2。

Therefore, MN/DM=(5√2)/(2√2)=5/2, that is, MN=(5/2)DM.

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