证明:延长BD,交AC于点N ∵AD⊥BN,AD平分∠BAN,AD=AD∴△ABD≌△AND∴AB=AN,BD=DN ∵M是BC的中点∴DM是△BCN的中位线∴DM=1/2CN =1/2(AC-AN)=1/2(AC-AB)
证明:延长BD,交AC于点E∵∠BAD=∠EAD,AD=AD,∠ADB=∠ADE=90°∴△ABD≌△AED∴AE=AB,BD=ED∵M是BC中点∴DM是△BCE的中位线∴MD=1/2CE=1/2(AC-AE)=1/2(AC-AB)