8. In the figure,AC and BD intersect at E. If AB = DC and AC = DB,prove that BE = CE.
19.In △ABC ,M is a point on BC.
(a) If AM is an angle bisector and an altitude of △ABC ,prove that △ABC is isosceles.
(b) If AM is a median and an altitude of △ABC, prove that △ABC is isosceles.
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2011-02-07 6:12 am
回答 (2)
2011-02-07 8:25 pm
✔ 最佳答案
BC=BC(common side)AB = DC(Given) and AC = DB(Given)
ABC~=DCB(SSS)
angleA=angleD(corr. angle,~=triangle)
AED=DEC(vert. opp. angle)
ABE~=DCE(AAS)
BE=CE(corr. side,~-triangle)
___________
AMR=AMC=90(Given)
RAM=CAM(Givem)
AM=AM(common side)
ARM~=ACM(ASA)
AR=AC(corr. side,~=triangle)
so,it is a isos. triangle
_________________
AM=AM(common side)
AMR=AMC=90(Given)
RM=MC(Given)
AMR~=AMC(SAS)
AR=AC(corr. side,~=triangle)
so,it is a isos. triangle
2011-02-07 9:30 am
Q8
Consider △ABC and △DCB,
AB = CD and AC = DB (given)
BC = BC (common)
Therefore, △ABC
圖片參考：http://upload.wikimedia.org/math/f/1/6/f16a7bd80bb5648a755eb58221d03546.png
△DCB (SSS)
Consider △ABE and △DCE
AB = CD (given)
angle AEB = angle DEC (vert.opp angles)
angle BAE = angle EDC (corr. angles,
圖片參考：http://upload.wikimedia.org/math/f/1/6/f16a7bd80bb5648a755eb58221d03546.png
△s)
Therefore, △ABE
圖片參考：http://upload.wikimedia.org/math/f/1/6/f16a7bd80bb5648a755eb58221d03546.png
△DCE (AAS)
AC = DB and AE = DE (corr. sides,
圖片參考：http://upload.wikimedia.org/math/f/1/6/f16a7bd80bb5648a755eb58221d03546.png
△s)
Therefore, BE = CE
Q19
(a)
angle BAM = angle CAM and angle BMA = angle CMA = 90 (given)
AM = AM (common)
Therefore, △ABM
圖片參考：http://upload.wikimedia.org/math/f/1/6/f16a7bd80bb5648a755eb58221d03546.png
△ACM (ASA)
AB = AC (corr. sides,
圖片參考：http://upload.wikimedia.org/math/f/1/6/f16a7bd80bb5648a755eb58221d03546.png
△s)
Therefore, △ABC is an isosceles triangle
(b)
BM = MC and angle BMA = angle CMA = 90 (given)
AM = AM (common)
Therefore, △ABM
圖片參考：http://upload.wikimedia.org/math/f/1/6/f16a7bd80bb5648a755eb58221d03546.png
△ACM (SAS)
AB = AC (corr. sides,
圖片參考：http://upload.wikimedia.org/math/f/1/6/f16a7bd80bb5648a755eb58221d03546.png
△s)
Therefore, △ABC is an isosceles triangle
Consider △ABC and △DCB,
AB = CD and AC = DB (given)
BC = BC (common)
Therefore, △ABC
圖片參考：http://upload.wikimedia.org/math/f/1/6/f16a7bd80bb5648a755eb58221d03546.png
△DCB (SSS)
Consider △ABE and △DCE
AB = CD (given)
angle AEB = angle DEC (vert.opp angles)
angle BAE = angle EDC (corr. angles,
圖片參考：http://upload.wikimedia.org/math/f/1/6/f16a7bd80bb5648a755eb58221d03546.png
△s)
Therefore, △ABE
圖片參考：http://upload.wikimedia.org/math/f/1/6/f16a7bd80bb5648a755eb58221d03546.png
△DCE (AAS)
AC = DB and AE = DE (corr. sides,
圖片參考：http://upload.wikimedia.org/math/f/1/6/f16a7bd80bb5648a755eb58221d03546.png
△s)
Therefore, BE = CE
Q19
(a)
angle BAM = angle CAM and angle BMA = angle CMA = 90 (given)
AM = AM (common)
Therefore, △ABM
圖片參考：http://upload.wikimedia.org/math/f/1/6/f16a7bd80bb5648a755eb58221d03546.png
△ACM (ASA)
AB = AC (corr. sides,
圖片參考：http://upload.wikimedia.org/math/f/1/6/f16a7bd80bb5648a755eb58221d03546.png
△s)
Therefore, △ABC is an isosceles triangle
(b)
BM = MC and angle BMA = angle CMA = 90 (given)
AM = AM (common)
Therefore, △ABM
圖片參考：http://upload.wikimedia.org/math/f/1/6/f16a7bd80bb5648a755eb58221d03546.png
△ACM (SAS)
AB = AC (corr. sides,
圖片參考：http://upload.wikimedia.org/math/f/1/6/f16a7bd80bb5648a755eb58221d03546.png
△s)
Therefore, △ABC is an isosceles triangle
收錄日期: 2021-04-13 17:48:33
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110206000051KK01171