`a,` Ta có: `AM` là cạnh chung.
`AB=AD`
`MB=MD`
`=>ΔAMB=ΔAMD(c.c.c)`
Dễ suy ra được: `AM⊥BD`
`b,` Ta có: `MK` là cạnh chung.
`∠BMK=∠DMK`
`=>ΔBKM=ΔDMK(c.g.c)`
`=>BK=KD`
c, Ta có:
`ΔAMB=ΔAMD`
`=>∠ABM=∠ADM`
`ΔAMK=ΔDMC`
`=>∠MBK=∠MDC`
`=>∠AMB+∠MBK=∠ADM+∠MDK`
`=>180^0-∠ABK=180^0-∠ADK`
`=>∠FBK=∠CDK`
Ta có: `FBK=∠CDK`
`BK=DK`
`BF=DC`
`=>ΔBKF=ΔDKC(c.g.c)`
`=>∠BKF=∠DKC`
Vì: `B,K,C` thẳng hàng nên:
`∠DKC+∠BKD=180^0`
Và: `∠BKF=∠DKC`
`=>∠BKF+∠BKD=180^0`
`=>F,K,D` thẳng hàng.
`=>Đpcm`