Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 671.
http://img838.imageshack.us/img838/3824/problem671.pngDraw triangle BDE as image of triangle BDA ( see picture)Note that BE=CD and (DCB)=(EBC)=4. alphaAnd DBCE is an isosceles trapezoid (BDE)=(BDA)= 180-7.alphaIn triangle ABD , x+180-7.alpha+3.alpha= 180 >> x=4. alphaIn triangle ABC , 7.alpha+4.alpha+4.alpha=180 >> alpha=10 and x=40Peter Tran
Okay, I have a really simple solution.At 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°.