Question

如图在△ABC中,AB=AC=1,点D,E在直线BC上运动,设BD=X,CE=Y

Answer 1

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-∠BAC

即∠ABC=∠DAE-∠BAC时

∵∠ABC=(180°-∠BAC)/2=90°-1/2∠BAC,∴当∠DAE-∠BAC=90°-1/2∠BAC

即∠DAE=90°+1/2∠BAC,M=90+N/2,Y与X的函数关系仍然成立