Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 671.
Thursday, September 22, 2011
Problem 671: Triangle, Cevian, Angles, Congruence
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ReplyDeleteDraw triangle BDE as image of triangle BDA ( see picture)
Note that BE=CD and (DCB)=(EBC)=4. alpha
And DBCE is an isosceles trapezoid
(BDE)=(BDA)= 180-7.alpha
In triangle ABD , x+180-7.alpha+3.alpha= 180 >> x=4. alpha
In triangle ABC , 7.alpha+4.alpha+4.alpha=180 >> alpha=10 and x=40
Peter Tran
Okay, I have a really simple solution.
ReplyDeleteAt first the problem seems to be awful but it can be taken down faster.
Let P be a point on BC so that DB = DP, hence <CDP = 3a and then ∆CDP = ∆ABD and x = 4a, hence a = 10° whereat x = 40°.
Very good simple solution.
DeleteFind E on DC such that < CBE = 3@
ReplyDeleteLet BD and BE extended meet the circle ABC at X and Y.
AX = CY so ACYX is an isoceles trapezoid.
Now < AEB = 7@ = < ABE
Hence AB = c = AE = CD
So AD = CE = b-c
Further < XAC = YCA
So tr.s AXD and CYE are congruent SAS.
< BDC = < BEA = 7@
Hence 7@ + 7@ + 4@ = 180 and therefore
@ = 10 and x = 40
Sumith Peiris
Moratuwa
Sri Lanka
[For easy typing, I use a instead of alpha]
ReplyDeletesin(180-14a)/BD=sin11a/AB
AB/BD=sin11a/sin14a--------(1)
sin7a/CD=sin4a/BD
CD/BD=sin7a/sin4a---------(2)
Since AB=CD, (1)=(2)
sin7a/sin4a=sin11a/sin14a
sin4asin11a=sin7asin14a
cos7a-cos15a=cos7a-cos21a
cos21a-cos15a=0
-2sin18asin3a=0
sin18a=0 or sin3a=0 (rej)
18a=180 [180 is the only suitable value]
a=10
x=180-14a=40