香港寵物小精靈村落論壇

標題: Similar Triangles [打印本頁]

作者: anthony 時間: 16/11/2005 07:36 PM
標題: Similar Triangles

In the above figure, AB,EF and DC are perpendicular to BC. Prove that
(a)△CEF∼△CAB
(b)△BEF∼△BDC
(c)1/a+1/b=1/c
我已經證明了(a)及(b)，但是不懂得證明(c)，希望大家可以幫幫忙，多謝。

[ Last edited by anthony on 2005-11-17 at 06:50 PM ]

作者: 宇文水靈 時間: 16/11/2005 10:33 PM
Since, △CEF∼△CAB,
      EF/AB=CF/CB
   i.e. c/a=CF/CB
         c/a=( BC-BF)/CB
         cBC= aBC-aBF
         aBF= BC(a-c)
      BF/BC=(a-c) /a------------------(1)
      △BEF∼△BDC
      EF/DC=BF/BC
         c/b=BF/BC--------------------(2)
From (1) & (2), c/b=BF/BC=(a-c)/a
                  c/b= (a-c)/a
                     ac=ab-cb-------------(*)
Divide  (*) by abc, 1/b= 1/c- 1/a
                  1/a+1/b=1/c

歡迎光臨香港寵物小精靈村落論壇 (https://archive.pokebeacon.com/discuz.hkpnve.net/)

香港寵物小精靈村落 論壇

香港寵物小精靈村落論壇