Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 856.
Sunday, February 10, 2013
Problem 856: Quadrilateral, Diagonals, Triangle, Angles, 30 degrees, Congruence, Auxiliary lines
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Draw DF perpendicular to AB and produce it to meet BC extended in F. It's easy to see that AFBD is a kite and Tr. AFD is equilateral. Further, Tr. DFC is isosceles with FC = FD. Tr.AFC is also isosceles with AF=FC which makes, /_FAC =(180-98)/2=41°. Further /_BAC=(41-30)=11° or x = 52 +11 = 63°_
ReplyDeleteVery nice solution Ajit .
DeletePeter Tran
Fie BM bisectoarea unghiului DBC ,M situat pe DC.=> triunghiul BMC isoscel cu masurile unghiurilor sale de 38°,71°,71° Paralela prin D la BM va intersecta dreapta BC in punctul N =>triunghiurile NBC ,NBD isoscele,NAD echilateral si NAC isoscel,de unde m(<ANC)=60+38=98 ,m(<ACN)=(180°-98°):2=41° in final x=m(<CEB)=180°-76°-41°= 63°
ReplyDeletetriunghiul BMC isoscel... De ce?
DeleteIn triunghiul DBC m(
Deletem(In triunghiul BMC m(triunghiul BMC isoscel
Thank you. Good solution.
ReplyDelete