Because ∠BAC and ∠BDC are BC arcs,

So ∠BAC=∠BDC

AE⊥CD,∠C+∠BAC=90

DF⊥AC,∠C+∠CDF=90

So ∠CDF=∠BAC=∠BDC

CD is the bisector of ∠BDF, and CD is the height on BM.

So △BDM is an isosceles triangle, and E is also the BM midline.

So BE = me.

(2) connect CF, AD and CM.

AB is the diameter and AB⊥CD, so AB is the middle vertical line of CD.

AC=AD，∠ADC=∠ACD

M is on AB, so CM=DM, ∠MDC=∠MCD.

So ∠ADC-∠MDC=∠ACD-∠MCD, that is ∠ADM=∠ACM.

DF=AC, so bad arc DF = bad arc AC.

Arc DC= arc DF- arc CF, arc AF= arc AC- arc CF.

So arc DC= arc AF, ∠ DFC = ∠ ACM = ∠ ADM

And ∠FMC=∠CMN.

So △FMC∽△CMN, Mn: cm = cm: MF.

CM？ =MN×MF, so DM =MN×MF

(3) connect CF, AD and CM.

AB is the diameter, AB⊥CD.

So arc AC= arc AD

AC=DF, so arc AC = arc DF.

So arc AD= arc DF

∠FCD is opposite to arc DF, ∞∠ACD is opposite to arc AD.

So ∠FCD=∠ACD.

∠ACD is the outer corner of △DCN, so ∠N=∠ACD-∠CDN.

CM=DM，∠CDN=∠MCD

∠FCM=∠FCD-∠MCD=∠FCD-∠CDN

So ∠N=∠FCM

And ∠NMC=∠CMF.

So △NMC∽△CMF, Mn: cm = cm: MF.

CM？ =MN×MF

Because DM=CM, DM? =MN×MF

(2) The conclusion is still valid.