Question

已知角b=角c=90度,ce=be,ae平分角dac,求证（1）de平分角adc.（2）如图2,

Answer 1

∵∠B=C=90°,即∠B+∠C=180°∴AB∥CD∵做AD中点F,连接EF,E是BC中点（CE=BE)∴EF∥AB∥CD∴∠BAE=∠FEA∵AE平分∠BAD∴∠BAE=∠FAE=∠FEA∴AF=EF=DF∴∠FDE=∠FED∵EF∥CD∴∠FED=∠CDE=∠FDF∴DE平分∠ADC