如图，在菱形ABCD中,E,F分别为BC,CD上的点,角B=角EAF=60度，角BAE=18度，求角CEF的度数

显示全部楼层 · 2010-11-20 00:39:30

解：连接AC∵四边形ABCD是菱形；∠B=60°∴△ABC和△ACD都是等边△∴∠BAE+∠EAC=60°∵∠EAF=60°=∠FAC+∠EAC∴∠FAC=∠BAE=18°∵△ABC和△ACD都是等边△∴AB=AC；∠B=∠ACF=60°∴△ABE≌△ACF∴AE=AF∵∠EAF=60°∴△AEF是等边△∴∠AEF=60°∵∠AEC=∠B+∠BAC=78°∴∠CEF=78°-60°=18°

千问 · 2010-11-20 00:39:30

解：连接AC∵四边形ABCD是菱形；∠B=60°∴△ABC和△ACD都是等边△∴∠BAE+∠EAC=60°∵∠EAF=60°=∠FAC+∠EAC∴∠FAC=∠BAE=18°∵△ABC和△A