有一三角形ABC,E為AB的中點,
則AC^2+BC^2=2(EC^2+AE^2)
答的方法有2種:
1. 話俾我知E條叫做咩定理,俾相關網頁我
2. 證明佢俾我睇.(F.6程度內我都睇得明)
話俾我知e條係咩定理or幫我證明佢
2007-09-17 12:58 am
回答 (2)
2007-09-17 1:28 am
✔ 最佳答案
畢氐定理+聯立就prove倒AE^2+EC^2=AC^2(畢氐定理)---1
AE^2+EC^2=BC^2(畢氐定理) (AE=EB)---2
1+2:
2(AE^2)+2(EC^2)=AC^2+BC^2
AC^2+BC^2=2(EC^2+AE^2)
自己畫幅圖會清楚D
2007-09-17 6:13 am
畢氐定理未必證到
因為冇話CE丄AB
我都唔知係咩定理
不過用cosine law 可以證到
AC^2=AE^2+EC^2-2(AE)(CE)cos角AEC
BC^2=EB^2+EC^2-2(EB)(CE)cos(180-角AEC)
BC^2=AE^2+EC^2+2(AE)(CE)cos角AEC...............SINCE AE=EB
So,
AC^2+BC^2=2(EC^2+AE^2)
唔明再問
因為冇話CE丄AB
我都唔知係咩定理
不過用cosine law 可以證到
AC^2=AE^2+EC^2-2(AE)(CE)cos角AEC
BC^2=EB^2+EC^2-2(EB)(CE)cos(180-角AEC)
BC^2=AE^2+EC^2+2(AE)(CE)cos角AEC...............SINCE AE=EB
So,
AC^2+BC^2=2(EC^2+AE^2)
唔明再問
收錄日期: 2021-04-13 13:31:51
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070916000051KK03348