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標題: 有數唔識做...@@" [打印本頁]

作者: oscar 時間: 30/3/2005 06:16 PM
標題: 有數唔識做...@@"
一條中三既數...
關於Deductive Geometry...
(b)題唔識...
請高人指點...
(第一次唔識做自己level既數...)

[ Last edited by oscar on 2005-3-30 at 06:19 PM ]

作者: 小耀 時間: 30/3/2005 06:34 PM

Originally posted by oscar at 2005-3-30 06:16 PM:
一條中三既數...
關於Deductive Geometry...
(b)題唔識...
請高人指點...
(第一次唔識做自己level既數...)

[ Last edited by oscar on 2005-3-30 at 06:19 PM ]

∵BE平分咗∠ABC,∴∠ABE=∠CBE
∠ABE=∠BEC(錯角,AB//EC)
∴∠CBE=∠BEC
DC=DC(公共邊)
BY(a) ∵BD=DE
∴∠BDC=∠DEC
∴△BDC≡△EDC(A.A.S.)
俾我紅咗果句嘢...總覺得:『好串』[E-Ton]

[ Last edited by 小耀 on 2005-3-30 at 07:32 PM ]

作者: oscar 時間: 30/3/2005 06:39 PM

Originally posted by 小耀 at 2005-3-30 18:34:

∵BE平分咗∠ABC,∴∠ABC=∠CBE
∠ABE=∠BEC(錯角,AB//EC)
∴∠CBE=∠BEC
DC=DC(公共邊)
BY(a) ∵BD=DE
∴BDC=DEC
∴△BDC≡△EDC

[ Last edited by 小耀 on 2005-3-30 a ...

你d steps又第1行已經錯...
∠ABC=∠CBE係錯...

作者: 艾 時間: 30/3/2005 06:48 PM

Originally posted by 小耀 at 2005-3-30 18:34:

∵BE平分咗∠ABC,∴∠ABC=∠CBE
∠ABE=∠BEC(錯角,AB//EC)
∴∠CBE=∠BEC
DC=DC(公共邊)
BY(a) ∵BD=DE
∴BDC=DEC
∴△BDC≡△EDC
俾我紅咗果句嘢...總覺得:『好 ...

∵BE BISECTS ∠ABC
∴∠ABD=∠CBD
∵∠ABE=∠BEC (ALT.∠S AB//EC)
∴∠CBD=∠DEC=∠ABD

上半應是這樣.....

作者: oscar 時間: 30/3/2005 06:49 PM

Originally posted by 颱風溫黛 at 2005-3-30 18:48:

∵BE BISECTS ∠ABC
∴∠ABD=∠CBD
∵∠ABE=∠BEC (ALT.∠S AB//EC)
∴∠CBD=∠DEC=∠ABD

上半應是這樣.....

我就係做黎做去都係S.S.A.

作者: 細薯 時間: 30/3/2005 06:58 PM

a)
since AB//EC and AB=EC.
ABCE is a parallelogram . (forget the reason)

therefore BD=DE (just remember bisect each other , forget the reason)

b)
angle BDA = angle BDC = angleEDC = 90 degrees (parallelogram ge property)

we now have

[quote]
BD = ED (from (a))
DC = DC (common)
angle BDC = angleEDC

therefore
triangle BDC & triangle EDC are congruent. (RHS)

[/quote]

sorry.........誤會左.....

[ Last edited by smallpotato226 on 2005-3-30 at 07:04 PM ]

作者: oscar 時間: 30/3/2005 07:06 PM
請大家盡量不要利用Properties of Quadrilaterals的知識來solve此題...

作者: 小耀 時間: 30/3/2005 07:09 PM

Originally posted by oscar at 2005-3-30 06:39 PM:

你d steps又第1行已經錯...
∠ABC=∠CBE係錯...

係咩...
乜原來錯架卅▽卅〣
-----------------------------------------
係喎...應該係∠ABE=∠CBE
原來自己打錯咗個英文字母,唔怪得知硬係覺得自己冇錯= ="
改番上面果度咯...咁咪A.A.S.囉...
冇錯呀...

[ Last edited by 小耀 on 2005-3-30 at 07:33 PM ]

作者: 小耀 時間: 30/3/2005 07:53 PM

Originally posted by 颱風溫黛 at 2005-3-30 06:48 PM:

∵BE BISECTS ∠ABC
∴∠ABD=∠CBD
∵∠ABE=∠BEC (ALT.∠S AB//EC)
∴∠CBD=∠DEC=∠ABD

上半應是這樣.....

相信這樣的一個小錯誤,在考試時,只能得到1分= ="
不過,打字打錯姐...希望考試果時,唔好寫錯就好了[E-Swe]

作者: 細薯 時間: 30/3/2005 07:54 PM
已和oscar 諗到啦
isos triangle...XD

作者: oscar 時間: 30/3/2005 07:56 PM
真係玩死我...
呢題數陪伴著我2粒鐘....

作者: 細薯 時間: 30/3/2005 07:57 PM

Originally posted by oscar at 2005-3-30 07:56 PM:
真係玩死我...
呢題數陪伴著我2粒鐘....

不
是 (b) part......XDXD

作者: oscar 時間: 30/3/2005 07:58 PM

Originally posted by smallpotato226 at 2005-3-30 19:57:

不
是 (b) part......XDXD

有時做數就是那麼無奈....

作者: 細薯 時間: 30/3/2005 08:01 PM

Originally posted by oscar at 2005-3-30 07:58 PM:

有時做數就是那麼無奈....

我都試過........

http://www.xanga.com/home.aspx?user=LWrabbit
Wednesday, March 23, 2005

Question

In triangle ABC, let a/sinA=b/sinB=c/sinC=k,where A,B and C are interior angles.

(a)show that a/sinA=a+b+c/sinA+sinB+sinC
(b)略
(c)略

my answer

i divide it into two part......
=======================
a+b+c=k(sinA+sinB+sinC)
=======================
sinA+sinB+sinC=(a+b+c)/2
=======================
therefore
a+b+c / sinA+sinB+sinC =k^2(sinA+sinB+sinC) / k(sinA+sinB+sinC)
=k
therefore = a /sin A

answer

a+b+c / sinA+sinB+sinC = k(sinA+sinB+sinC) / sinA+sinB+sinC = k = sin A

.................

[ Last edited by smallpotato226 on 2005-3-30 at 08:07 PM ]

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