Question

如图,直角三角形abcd中,ad平行bc,角A等于90度,DE垂直DC交AB于E,DF平分角EDC交BC于F,连接EF&

Answer 1

过D作DG⊥BC交BC于G.

∵AD∥BG、AD⊥AB、DG⊥BG,∴ABGD是矩形,又AB＝AD,∴矩形ABGD是正方形,

∴AD＝GD、∠ADG＝∠DGC＝90°,∴∠ADE＋∠EDG＝∠EDG＋∠GDC,

∴∠ADE＝∠GDC,∴△ADE≌△GDC,∴∠DE＝DC,又DF＝DF、∠EDF＝∠CDF,

∴△EDF≌△CDF,∴EF＝CF.