连接BE.
BE是RT△ABC斜边上的中线,可得:BE = CE ;
所以,∠AEB = 2∠ACB = 60°,∠BED = ∠AED+∠AEB = 150°;
DE是RT△ADC斜边上的中线,可得:DE = CE ;
所以,CE = DE ,∠BDE = (1/2)(180°-∠BED) = 15°.
连接BE.
BE是RT△ABC斜边上的中线,可得:BE = CE ;
所以,∠AEB = 2∠ACB = 60°,∠BED = ∠AED+∠AEB = 150°;
DE是RT△ADC斜边上的中线,可得:DE = CE ;
所以,CE = DE ,∠BDE = (1/2)(180°-∠BED) = 15°.