數學題 triangle+inscribed circle

圖片參考：http://img6.glowfoto.com/images/2010/07/05-0650007518L.gif


1) △AGD ~ △ADE ,

so AG/AD = AD/AE

AD^2 = AG * AE

x^2 = AG * AE

;;;;

△ADF / (AG*AE)

= (x^2)sin60° / (AG*AE)

= sin60° = √3 / 2 for A

2)

BD = BE
and CF = CE

so BC = BE + CE = BD + CF = 4 + 2 = 6 for B

In △ABC , by cosine formula ,

cos60° = (AB^2 + AC^2 - BC^2)/(2AB AC)

1/2 = (x+4)^2 + (x+2)^2 - 6^2) / [2 (x+4)(x+2)]

(x+4)(x+2) = (x+4)^2 + (x+2)^2 - 6^2

x^2 + 6x + 8 = x^2 + 8x + 16 + x^2 + 4x + 4 - 36

x^2 + 6x - 24 = 0 for C , D

x = {- 6 +/- √[6^2 - 4(-24)]} / 2

x = - 3 + √33 for E or - 3 - √33 (rejected)

點解 △AGD ~ △ADE??