[問題]S3 Maths - Deductive Geometry

oscar

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~

JOHNLAM17

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.........

細薯

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 ]

oscar

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...

JOHNLAM17

冇人理我o既...
我個answer長左d je...=.=

oscar

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...

抹茶可樂

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 ]

JOHNLAM17

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

oscar

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?

JOHNLAM17

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.