## Wednesday, November 30, 2011

### Problem 694: Triangle, Angles, 30 Degrees, Auxiliary Construction, Mind Map, Polya

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

Click the figure below to see the complete problem 694.

1. Draw a perpendicular BF from B to AC intersecting CD in E, join AE. Since Tr ABC is isosceles, EA =EC and therefore /_EAC=30 hence /_DAE=18 or AD is angle bisector of /_BAE. Now /_AEF=60=/AED and thus /_DEB=60 so ED bisects /_AEB which means BD bisects /_ABE. So /_x =12

2. Here is the generalization of this problem

http://www.fmat.cl/index.php?s=&showtopic=74752&view=findpost&p=567874

3. I still don't understand...

4. AB = BC, let BE be the perpendicular bisector of AC meeting DC at E. By computing angles we see that D is the in centre of Tr. ABE hence x = 12

Sumith Peiris
Moratuwa
Sri Lanka

5. Problem 694
Forming equilateral ADE (point E is lower thanBC ),then <AED=60=2.30=2<ACD so E is circumcenter of triangle ADC,so AE=AD=DE=EC. But triangle ABE= triangle BCE,then
AB=BE and <ABE=48/2=24.But triangle ABD=triangle DBE (AB=BE,AD=DE, BD=BD),
And <ABD=x=24/2=12.