延长CM至H,使CM=MH,连接FH、BH、CM、BM,延长CD,与BF相交于I
∵MF=MD CM=HM ∠CMD=∠HMF
∴△CMD≌△HMF
HF=CD=AC
∠HFJ=180°-∠JHF-∠HJF
∠HJF=∠IJC ∠JHF=∠DCM
∠BIC=∠IJC+∠DCM
四边形ABIC中∠ABI=∠ACI=RT∠
∠BAC=360°-∠ABI-∠ACI-∠BIC=180°-∠BIC=180°-∠IJC-∠DCM=180°-∠JHF-∠HJF=∠HFB
∴△ABC≌△FBH
∠HBF=∠ABC
∠CBH=∠HBF+∠CBF=∠ABC+∠CBF=90°
BC⊥BH
N是BC中点,M是HC中点
MN‖BH
BC⊥MN