Question

如圖，分別以Rt△ABC的直角邊AC及斜邊AB向外作等邊△ACD、等邊△ABE.已知∠BAC＝30°，EF⊥AB，垂足為F，連結DF.

(1)試證明AC＝EF.
(2)求證：四邊形ADFE是平行四邊形．

Answer 1

見解析

證明：(1)∵等邊△ABE中，EF⊥AB
∴EF平分∠AEB，∴∠AEF＝∠AEB＝30°
∵∠BAC＝30°，∴∠AEF＝∠BAC
又∵∠AFE＝∠ACB＝90°，AE＝AB
∴△ABC≌△EAF　∴AC＝EF
(2)∵等邊△ACD中，∠DAC＝60°
而∠CAB＝30°，∴∠DAF＝90°＝∠AFE
∴AD∥EF　又∵AD＝AC，AC＝EF
∴AD＝EF.∴四邊形ADFE是平行四邊形．