Giải thích các bước giải:
a.Xét $\Delta ABH,\Delta ACH$ có:
Chung $AH$
$HB=HC$ vì $H$ là trung điểm $BC$
$AB=AC$
$\to\Delta ABH=\Delta ACH(c.c.c)$
b.Xét $\Delta AHB,\Delta MHC$ có:
$HB=HC$
$\widehat{AHB}=\widehat{MHC}$
$HA=HM$
$\to\Delta AHB=\Delta MHC(c.g.c)$
$\to \widehat{ABH}=\widehat{HCM}\to AB//CM$
3.Xét $\Delta BKC,\Delta BKD$ có:
Chung $BK$
$\widehat{BKC}=\widehat{BKD}=90^o$ vì $BK\perp AC$
$KC=KD$
$\to\Delta BKC=\Delta BKD(c.g.c)$
$\to\widehat{CBK}=\widehat{DBK}$
$\to BK$ là phân giác $\widehat{CBD}$
4.Từ câu a
$\to\widehat{ABH}=\widehat{ACH}$
Mặt khác theo câu c $\to\widehat{BDK}=\widehat{BCK}, BC=BD$
$\to\widehat{ABH}=\widehat{BDK}$
$\to 180^o-\widehat{ABH}=180^o-\widehat{BDK}$
$\to \widehat{EBC}=\widehat{ADB}$
Xét $\Delta ABD,\Delta ECB$ có:
$BC=BD$
$\widehat{ADB}=\widehat{CBE}$
$AD=BE$
$\to\Delta ABD=\Delta ECB(c.g.c)$
$\to AB=CE$
Mà $AB=AC\to CE=CA$
