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
