∠BMC=90°求详解两邻边AB与AD的比是1:2,M是AD的中点∴AB=AM∴∠ABM=∠AMB∵AD∥BC∴∠AMB=∠CBM＝∠ABM=½∠ABC同理∠BCM＝∠DCM=½∠BCD∵AB∥CD∴∠ABC＋∠BCD=180°∴∠MBC＋∠BCM=½×180°＝90°∴∠BMC=90°