## Friday, February 12, 2010

### Geometry Problem 425: Quadrilateral, Triangle, Angles, 10, 20, 50 degrees, Congruence

Proposed Problem
Click the figure below to see the complete problem 425 about Quadrilateral, Triangle, Angles, 10, 20, 50 degrees, Congruence.

Complete Problem 425

Level: High School, SAT Prep, College geometry

1. http://geometri-problemleri.blogspot.com/2010/02/problem-73-ve-cozumu.html

2. Here is a diferent solution from the previous post:

http://ricardosandoval.wordpress.com/2010/04/11/solucao-em-ingles-para-um-problema-da-internet/

3. Another solution that constructs only one new point:

http://ricardosandoval.wordpress.com/2010/04/11/solucao-em-ingles-para-um-problema-da-internet/

And doesn't uses circle properties as in:

http://amenazatriangular.blogspot.com/

4. solution
http://cebirci.net/forum/ozel-geometri-sorulari/10-20-50-x.html#277
by hanzala

5. Drawing circumcircle of BCD turns the problem into a cevian model 20,10,30,20,120-x,x-20.
General solution of that kind of cevian model listed here: http://geomania.org/forum/fantezi-geometri/ucgen-icerisinde-p-noktasi-model-4-3/ but in Turkish. But you can follow via pictures.
Remember that in a cevian model, some angles can be interchanged.

6. Draw perpendiculars BX from B to DC and DY from D to AB

< DAC = 20 and < BDY = 40 = < BDC hence Tr.s BDY and BDX are congruent ASA

Tr. ADY is 30-60-90 and so DY = DA /2 = DC/2 = DX. Hence DX = XC and so Tr. BDC is isoceles and x = 100

Sumith Peiris
Moratuwa
Sri Lanka

7. Problem 425
Dialing the BDE equilateral triangle, so < DEB = 60 = 2.30 = 2.<DAB, therefore AE = ED = DB = EB , <EAB = 10=<EBA.But triangle AED=triangle CDB( AD=DC,ED=DB,<ADE=40=<CDB ).So