Question

AB是直径,弦AC=12,BC=16,∠ABC的角平分线交圆O于点D,求AD的长?

Answer 1

∵AB是直径,∴∠C=∠D=90°,

∴AB=√(AC^2+BC^2)=20,

∴cos∠ABC=BC/AB=4/5,

∵AD平分∠ABC,

∴∠ABD=1/2∠ABC,

∴sin∠ABD=√［(1-cos∠ABC)/2]=√(1/10)=√10/10,

∴AD/AB=√10/10,

AD=2√10.