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

Labels:
30 degrees,
angle,
congruence,
diagonal,
quadrilateral,
triangle

Subscribe to:
Post Comments (Atom)

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