## 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.

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°_

1. Very nice solution Ajit .
Peter Tran

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°

1. triunghiul BMC isoscel... De ce?

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

3. Thank you. Good solution.