Question

问一道数学题,急如图在△ABC中,AB=AC,∠C=72°,⊙O过AB两点且与BC切于B,与AC交于D,连结BD,若BC

Answer 1

AB = AC,∠C =72,则 ∠A = 36 ,∠ABC =72.

由圆周角定理：∠CBD = ∠A = 36 ,所以 ∠ABD = 36

所以 三角形ABD为等腰三角形,三角形BCD也为等腰三角形.

所以 BC = BD = AD = 1 ,设 CD =X .则AC = X + 1

由相似三角形定理得,AC:BC = BD :DC 得 X = (根号5 - 1)/2

所以 AC = 1 + (根号5 - 1)/2 = (根号5 + 1)/2