Question

如图,已知△ABC中,AB=AC=√5,BC=4,点O在BC边上运动,以O为圆心,OA为半径的圆与边AB交于点D（点A除

Answer 1

AB=AC=√5,BC=4=>cos∠ABC=(BC/2)/AB=2/√5

OB=x,=>

OA^2=AB^2+OB^2-2AB*OB*cos∠ABC=5+x^2-4x

=>cos∠OAB=(AB^2+OA^2-OB^2)/(2AB*OA)

AD=2OA*cos∠OAB=(AB^2+OA^2-OB^2)/AB=(5+5+x^2-4x-x^2)/√5=(10-4x)/√5=y

当10-4x0时,y->2√5.

所以0