假设在等腰三角形ABC中AB=AC,CD垂直AB.当D点靠近A点时,设AB=AC=X,CD=Z,BD=Y则2X-Y=12,Y+Z=18,2X+Z=30,在直角三角形ADC与BDC中AC^2-AD^2=CD^2=BC^2-BD^2,所以X^2-(12-X)^2=Z^2-(18-Z)^2,解得X=(3Z-15)/2,代入2X+Z=30中,解得X=75/8,Z=45/4,当D靠近B点时,2X-Y=18,Y+Z=12,2X+Z=30,X^2-(18-X)^2=Z^2-(12-Z)^2,Z=(3X-15)/2,代入,X=3,Y=-12,Z=24,而Y大于0,所以舍去.所以AB=AC=75/8,BC=45/4