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°.