[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~

細薯

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 .......

oscar

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！

細薯

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...><

oscar

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~

細薯

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~)

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.

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 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

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 ]