Sunday, February 10, 2013

Problem 856: Quadrilateral, Diagonals, Triangle, Angles, 30 degrees, Congruence, Auxiliary lines

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 856.

Online Geometry Problem 856: Quadrilateral, Diagonals, Triangle, Angles, 30 degrees, Congruence, Auxiliary lines

6 comments:

  1. 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°_

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  2. prof radu ion Sc Bozioru, BuzauFebruary 11, 2013 at 11:55 AM

    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°

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    Replies
    1. triunghiul BMC isoscel... De ce?

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    2. In triunghiul DBC m(
      m(In triunghiul BMC m(triunghiul BMC isoscel

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