如图,在△ABC中,AB=AC=1如图,在△ABC中,AB=AC=1,点D,E在直线BC上运动...答:解:1.∵∠BAC=30°,AB=AC,∴∠ABC=∠ACB=75° ∠ABC是△ABD外角,∴∠D+∠DAB=75°.∠CAE+∠DAB=∠DAE-∠BAC=75°.∴∠D=∠CAE.∠ABD=180°-∠ABC,∠ACE=180°-∠ACB.∴∠ABD=∠ACE △ABD∽△ACE.BD/AB=AC/CE,X/1=1/Y.所以Y=1/X,2.当∠D+∠DAB=∠CAE+∠DAB=∠DAE...