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

Posted by Antonio Gutierrez at 1:17 PM
Labels: , , , , ,

6 comments:

  1. AjitFebruary 10, 2013 at 9:43 PM

    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°_

    ReplyDelete
  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°

    ReplyDelete
    Replies
    1. AnonymousFebruary 12, 2013 at 7:28 AM

      triunghiul BMC isoscel... De ce?

      Delete
    2. prof radu ionFebruary 14, 2013 at 5:42 AM

      In triunghiul DBC m(
      m(In triunghiul BMC m(triunghiul BMC isoscel

      Delete
  3. UnknownJuly 26, 2013 at 4:20 AM

    Thank you. Good solution.

    ReplyDelete

Share your solution or comment below! Your input is valuable and may be shared with the community.

Subscribe to: Post Comments (Atom)