夏日香氣

圖中,E是BD上的一點,使CE=DE
3(弧AB)=6(弧BC)=6(弧CD)=(弧AD)
求角ACE
答案係90度,求步驟~()

[ 本帖最後由 夏日香氣 於 2010-10-17 04:49 PM 編輯 ]

KT6491

First find the relations among arcs AB, BC, CD, and AD in ratio,
i.e. arc AB: arc BC: arc CD: arc AD = 2:1:1:6.
Then apply "arcs proportional to angles at circumference" to find the relations among the angles in the triangles.
I believe you can do the subsequent steps. Be confident!

夏日香氣

引用:

原帖由 KT6491 於 2010-10-17 05:08 PM 發表
First find the relations among arcs AB, BC, CD, and AD in ratio,
i.e. arc AB: arc BC: arc CD: arc AD = 2:1:1:6.
Then apply "arcs proportional to angles at circumference" to find the relations among  ...

我都係用你所指既方法,但係計唔到~

firoazen

Hints: ACE=ACD-ECD, where ECD=EDC=BAC

夏日香氣

引用:

原帖由 firoazen 於 2010-10-17 05:19 PM 發表
Hints: ACE=ACD-ECD, where ECD=EDC=BAC

呢d資料我都知,可唔可以全題解俾我~

firoazen

ACD=ABC=180x[6/(2+1+1+6)]=108
ECD=EDC=180x[1/(2+1+1+6)]=18
Hence,
ACE=ACD-ECD=108-18=90

夏日香氣

引用:

原帖由 firoazen 於 2010-10-17 05:25 PM 發表
ACD=ABC=180x[6/(2+1+1+6)]=108
ECD=EDC=180x[1/(2+1+1+6)]=18
Hence,
ACE=ACD-ECD=108-18=90

原來我誤會左,以為弧AD=弧AB+弧BC+弧CD~
禁幾時先知弧係指優弧(大於弧一半)定劣孤(小於弧一半)?
同埋點解要用180x<~