Question

已知等腰三角形ABC中,顶角角A=36度,BD为角ABC的平分线与AC交于D,则AD/AC=?

Answer 1

设AC=1,AD=X>0,∠CBD=∠ABD=∠A=36°1/X=AC/BC=BC/CD=BD/CD=AD/(AC-CD)=X/(1-X),X?+X-1=0AD/AC=X/1=X=[-1+√(1?+4×1×1)]/(2×1)=(√5-1)/2≈0.618黄金分割