如图，四边形ABCD，∠A=130°，点D在AB，AC的垂直平分线上，则∠BDC=

显示全部楼层 · 2013-4-12 21:50:19

解：如图作DE⊥AB，DF⊥AC,连接DA∵点D在AB，AC的垂直平分线上∴∠B=∠DAE，∠C=∠DAF则 ∠B+∠C=∠DAE+∠DAF=∠A=130°∴∠BDC=360°-∠A-∠-B∠C=360°-130°-130°=100°. ...

千问 · 2013-4-12 21:50:19

连接AD点D在AB的垂直平分线上所以 DA = DB点D在AC的垂直平分线上所以 DA = DC得2等腰三角形ADB, ADC∠B =∠ DAB∠C = ∠DAC∠B+∠C = ∠DAB + ∠DAC = 130°∠D = 360° - 130° - ∠B -∠C =360 °-130° -130° =...