✔ 最佳答案
a) ang. CAB = ang. EDB ( corr angles, Ac parallel DE)
ang. EDB = ang. EFB ( angles in the same seg.)
therefore
ang. CAB = ang. EFB
because
ang. CAB = ang. EFB
therefore A,F,B,C are concyclic
( converse of angles in the same seg)
b) because M is the mid. pt of AC and A,F,B,C are concyclic
therefore
M is the centre of the circle passing A,F,B,C
therefore
MA, MF, MB, MC are the radii of the circle AFBC
then
MF=MB
c) ang. MAD + ang.BAF = ang. MAF
ang. MAF = ang. MFA (bass angles, isos. triangle)
ang. MCE + angFCB = ang. MCB
ang. MCB = ang. MBC (bass angles, isos. triangle)
ang. BAF = ang. FCB ( angles in the same seg.)
ang. MFA = ang. MBF (angles in alt. seg.)
ang. MBC = ang. MFB (angles in alt. seg.)
ang. MBF = ang. MFB (bass angles, isos. triangle)
therefore
ang. MAD + ang.BAF = ang. MCE + angFCB
ang. MAD + ang.FCB = ang. MCE + angFCB
ang. MAD = ang. MCE
ang. MAD = ang. BFC
ang. MCE = ang. BFC
because
ang. MCE = ang. BFC
therefore
AC parallel to FB (alt. angles, eq. angles)
because
AC parallel to FB and DE respectively,,
therefore
DE parallel to FB
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