香港寵物小精靈村落 論壇

標題: [問題]S3 Maths - Deductive Geometry [打印本頁]
作者: oscar    時間: 14/4/2005 08:00 PM
標題: [問題]S3 Maths - Deductive Geometry
Questions:
From the diagram, ABCD is a quadrilateral. AC and BD intercepts at point E. Given that AD=DC=CB and AB//CD. Prove that Triangle ABE is an isosceles triangle.
請用英文作答！
Thx~
作者: 細薯    時間: 14/4/2005 08:32 PM
angle CDB = angle DBA (alt.angles,CD//AB)
angle CDB = angle CBD (base angles,isos.triangle)

angle DCA = angle CAB (alt.angles,CD//AB)
angle DCA = angle DAC (base angles,isos.triangle)

angle AEB = angle CED (vert.opp.angles)

i just know that triangle AEB ~ triangle CED .......
作者: oscar    時間: 14/4/2005 08:35 PM
Originally posted by smallpotato226 at 2005-4-14 20:32:
angle CDB = angle DBA (alt.angles,CD//AB)
angle CDB = angle CBD (base angles,isos.triangle)

angle DCA = angle CAB (alt.angles,CD//AB)
angle DCA = angle DAC (base angles,isos.triangle)

angle ...

那麼我也知道...
這題並不容易！
連我的數學老師都不懂這題！
作者: 細薯    時間: 14/4/2005 08:38 PM
Originally posted by oscar at 2005-4-14 08:35 PM:

那麼我也知道...
這題並不容易！
連我的數學老師都不懂這題！

嗯
那等我抄低tomorrow返school問人~

ps:如果用得Form 4 knowledge 就好易...><
作者: oscar    時間: 14/4/2005 08:39 PM
Originally posted by smallpotato226 at 2005-4-14 20:38:

嗯
那等我抄低tomorrow返school問人~

ps:如果用得Form 4 knowledge 就好易...><

我只知道用Sine Law做不到~
作者: 細薯    時間: 14/4/2005 08:44 PM
Originally posted by oscar at 2005-4-14 08:39 PM:

我只知道用Sine Law做不到~

用cylic quad. ~~XDXD
(converse of angles in the same segment~)
作者: JOHNLAM17    時間: 14/4/2005 09:43 PM
Originally posted by oscar at 2005-4-14 08:00 PM:
Questions:
From the diagram, ABCD is a quadrilateral. AC and BD intercepts at point E. Given that AD=DC=CB and AB//CD. Prove that Triangle ABE is an isosceles triangle.
請用英文作答！
Thx~


In Triangle ADC and Triangle BCD
AD = BC (given)
跟住自己將個quadrilateral 畫上去
變左做triangle<---知唔知我講咩
then I call the 頂點of triangle is "F"
Angle FDC = Angle FAB (corr. angles, AB//CD)
Angle FCD = Angle FBA (corr. angles, AB//CD)
Angle ADC = 180 - Angle FDC (adj. angles on st. line)
Angle BCD = 180 - Angle FCD (adj. angles on st. line)
Because Triangle FAB is an isosceles triangle,<---DA = CB,so FA = FB
so Angle FAB =Angle FBA = Angle FDC = Angle FCD
So Angle ADC = Angle BCD
DC = CD (common side)
So triangle ADC = Triangle BCD (S.A.S.)
So Angle BDC = Angle ACD (corr. angles, = triangle)
So Angle BDC = Angle DBA (alt. Angles, AB//CD)
     Angle ACD = Angle CAB (alt. angles, AB//CD)
So Angle DBA = Angle CAB
So Triangle ABE is an isosceles triangle.
作者: oscar    時間: 14/4/2005 09:51 PM
Originally posted by JOHNLAM17 at 2005-4-14 21:43:
In Triangle ADC and Triangle BCD
AD = BC (given)
跟住自己將個quadrilateral 畫上去
變左做triangle<---知唔知我講咩
then I call the 頂點of triangle is "F"
Angle FDC = Angle FAB (corr. angles, AB//CD)
Angle FCD = Angle FBA (corr. angles, AB//CD)
Angle ADC = 180 - Angle FDC (adj. angles on st. line)
Angle BCD = 180 - Angle FCD (adj. angles on st. line)
Because Triangle FAB is an isosceles triangle,<---DA = CB,so FA = FB
so Angle FAB =Angle FBA = Angle FDC = Angle FCD
So Angle ADC = Angle BCD
DC = CD (common side)
So triangle ADC = Triangle BCD (S.A.S.)
So Angle BDC = Angle ACD (corr. angles, = triangle)
So Angle BDC = Angle DBA (alt. Angles, AB//CD)
     Angle ACD = Angle CAB (alt. angles, AB//CD)
So Angle DBA = Angle CAB
So Triangle ABE is an isosceles triangle.

紅色o個一步錯！
你點樣prove DA=CE, so FA=FB?
作者: JOHNLAM17    時間: 14/4/2005 09:56 PM
Originally posted by oscar at 2005-4-14 09:51 PM:

紅色o個一步錯！
你點樣prove DA=CE, so FA=FB?

Because DC//AB
So FD/DA = FC/CB
Because DA = CB(given)
so FD = DC (equal ratios theroem)
so FA =FB
作者: 抹茶可樂    時間: 14/4/2005 10:57 PM
ADP≡BCQ(RHS)
∠DAP=∠CBQ
ADB≡BCA(SAS)
AC=BD
ADC≡BCD(SSS)
∠DCE=∠CDE
∠EBA=∠DAB
AEB is isoscles

*only key steps are written

[ Last edited by 落雷 on 2005-4-14 at 11:05 PM ]
作者: oscar    時間: 14/4/2005 11:54 PM
Originally posted by 落雷 at 2005-4-14 22:57:
ADP≡BCQ(RHS)
∠DAP=∠CBQ
ADB≡BCA(SAS)
AC=BD
ADC≡BCD(SSS)
∠DCE=∠CDE
∠EBA=∠DAB
AEB is isoscles

*only key steps are written

[ Last edited by 落雷 on 2005-4-14 at 11:05 PM ]

哈哈！你的方法與我校另一位數學老師的方法一樣！
好野！等我仲諗左成3日...
作者: JOHNLAM17    時間: 15/4/2005 07:24 PM
冇人理我o既...
我個answer長左d je...=.=
作者: oscar    時間: 15/4/2005 07:57 PM
Originally posted by JOHNLAM17 at 2005-4-15 19:24:
冇人理我o既...
我個answer長左d je...=.=

如果以中三的程度去做，你的方法是太複雜...
作者: 細薯    時間: 16/4/2005 12:45 PM
Originally posted by 落雷 at 2005-4-14 10:57 PM:
ADP≡BCQ(RHS)
∠DAP=∠CBQ
ADB≡BCA(SAS)
AC=BD
ADC≡BCD(SSS)
∠DCE=∠CDE
∠EBA=∠DAB
AEB is isoscles

*only key steps are written

[ Last edited by 落雷 on 2005-4-14 at 11:05 PM ]


To 落雷：
prove到 red part is enough.
ADB≡BCA(SAS)
∠DBA = ∠CAB

[ Last edited by smallpotato226 on 2005-4-16 at 12:47 PM ]
作者: JOHNLAM17    時間: 16/4/2005 02:25 PM
Originally posted by oscar at 2005-4-15 07:57 PM:

如果以中三的程度去做，你的方法是太複雜...

但係我基本上只係prove兩個triangle congrent.........



歡迎光臨 香港寵物小精靈村落 論壇 (https://proxy.archiver.hkpnve.pokebeacon.com/)Powered by Discuz! X3.2