Giải thích các bước giải:

a.Xét $\Delta AMB,\Delta AMD$ có:

Chung $AM$

$AB=AD$

$MB=MD$

$\to \Delta AMB=\Delta AMD(c.c.c)$

b.Từ câu a $\to \widehat{AMB}=\widehat{AMD}$

Mà $\widehat{AMB}+\widehat{AMD}=\widehat{BMD}=180^o\to \widehat{AMB}=\widehat{AMD}=90^o$

$\to AM\perp BD$

c.Từ câu a$\to \widehat{BAM}=\widehat{DAM}\to \widehat{BAK}=\widehat{DAK}$

Xét $\Delta ABK,\Delta DAK$ có:

Chung $AK$

$\widehat{BAK}=\widehat{DAK}$

$AB=AD$

$\to \Delta ABK=\Delta ADK(c.g.c)$

d.Từ câu c $\to \widehat{ABK}=\widehat{ADK}$

$\to 180^o-\widehat{ABK}=180^o-\widehat{ADK}$

$\to \widehat{FBK}=\widehat{KDC}$

Mặt khác $KB=KD$

Xét $\Delta KBF,\Delta KDC$ có:

$KB=KD$

$\widehat{KBF}=\widehat{KDC}$

$BF=CD$

$\to \Delta KBF=\Delta KDC(c.g.c)$

$\to \widehat{BKF}=\widehat{DKC}$

$\to \widehat{DKF}=\widehat{DKB}+\widehat{BKF}=\widehat{DKB}+\widehat{DKC}=\widehat{BKC}=180^o$

$\to F, K, D$ thẳng hàng