发布时间:2021-02-22 23:41:23
如图,分别以Rt△ABC的直角边AC及斜边AB向外作等边△ACD、等边△ABE.已知∠BAC=30º,EF⊥AB,垂足为F,连结DF.
(1)求证:AC=EF;
(2)求证:四边形ADFE是平行四边形.
【解析】由等边△ABE和Rt△ABC,求得Rt△ABC∽Rt△EAF,即可得AC=EF,由等边三角形的性质得出∠BDF=30°,从而证得△DBF≌△EFA,则AE=DF,再由FE=AB,得出四边形ADFE为平行四边形
(1)∵在等边△ABE中,EF⊥AB,
∴AF= AE= AB,
又∵Rt△ABC,∠BAC=30º,
∴BC=AB,
∴BC=AF
∴Rt△ABC∽Rt△EAF(AAS)
即AC=EF
(2)因为EF⊥AB,∴,∠AFE=90
∵△ACD是等边三角形,∴∠DAC=60,∴∠DAB=90
∵∠AFE=∠DAB,∴AD//EF
∵∠BAC=30,∴CB=AB
∵EF⊥AB,∴AF=AB=CB
∵AF=CB.AD=AC,∠DAB=∠ACB=90
∴Rt△ABC∽Rt△DFA
∴∠ADF=∠CAB=30,
∵∠DAB+∠BAE=90+60=150
∴∠ADF+∠DAE=180
∴AE//DF
∴四边形ADFE是平行四边形