图是纵向交流,CF垂直于AB于点E,F,BE和CF相交于点D,DE = DF,连接到AD证明:1,∠FAD = 2,BD = CD EAD

问题描述:

图是纵向交流,CF垂直于AB于点E,F,BE和CF相交于点D,DE = DF,连接到AD证明:1,∠FAD = 2,BD = CD EAD

∵∠AFD = 90?
,精益所以ED,DF = DE,AD市民,∴△AFD≌△AED,∴∠FAD =∠EAD,∵∠DFB =∠DEC,DF =
DE,∠教育局=∠EDC,∴△DFB≌△DEC,∴ BD = CD.