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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 694.
Draw a perpendicular BF from B to AC intersecting CD in E, join AE. Since Tr ABC is isosceles, EA =EC and therefore /_EAC=30 hence /_DAE=18 or AD is angle bisector of /_BAE. Now /_AEF=60=/AED and thus /_DEB=60 so ED bisects /_AEB which means BD bisects /_ABE. So /_x =12
Here is the generalization of this problemhttp://www.fmat.cl/index.php?s=&showtopic=74752&view=findpost&p=567874
I still don't understand...
AB = BC, let BE be the perpendicular bisector of AC meeting DC at E. By computing angles we see that D is the in centre of Tr. ABE hence x = 12Sumith PeirisMoratuwaSri Lanka
Problem 694Forming equilateral ADE (point E is lower thanBC ),then <AED=60=2.30=2<ACD so E is circumcenter of triangle ADC,so AE=AD=DE=EC. But triangle ABE= triangle BCE,then AB=BE and <ABE=48/2=24.But triangle ABD=triangle DBE (AB=BE,AD=DE, BD=BD),And <ABD=x=24/2=12.