國中數學幾何問題~急!!!

圖片參考：http://img5.imageshack.us/img5/484/40178066.png



1) 設 EF = EC = x ,

則 AE = AF + FE = 4 + x

DE = DC - EC = 4 - x

勾股定理 :

AE^2 = AD^2 + DE^2

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

8(2x) = 16

x = 1

△ADE = (1/2) DE * AD = (1/2) (4 - 1) (4) = 6

△ADF = (AF/AE) * △ADE = (4/(4+1)) * 6 = 24/5


2)


圖片參考：http://img196.imageshack.us/img196/2126/123cez.png

BF + FC = 10 ...........(1)
;
AD = AB + BD = 6 + BF
;
AE = AC + CE = 8 + FC

因 AD = AE ,

故 BF + 6 = FC + 8 .........(2)

(1) - (2) :

FC - 6 = 2 - FC
FC = 4

勾股定理 :

BE^2 = AB^2 + AE^2

BE^2 = 6^2 + (AC + CE)^2

BE^2 = 36 + (AC + FC)^2

BE^2 = 36 + (8 + 4)^2

BE^2 = 180

BE = √180

BE = 6√5




2010-10-27 23:01:29 補充：
一個必須說明的重要事項 :

10^2 = 6^2 + 8^2

BC^2 = AB^2 + AC^2

由勾股逆定理 , ㄥBAC 是直角。

2010-10-27 23:30:57 補充：
不用客氣^^....