## Wednesday, December 12, 2012

### Problem 830: Quadrilateral, Triangle, Angles, 30 degrees, Congruence

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

Click the figure below to see the complete problem 830.

1. http://img577.imageshack.us/img577/4899/problem830.png

See drawing per attached sketch
Extend BC and from D draw DF ⊥to BC
We have ∠BDF=60 and DF=1/2. BD
In isosceles triangle ACD draw altitude CE
CE and CF are tangent lines from point C to circle centered D with radius=DE=DF ( see Sketch)
∠EDF=60+34=94
∠FCD=90-1/2. 94=43
So in triangle BCD, external angle FCD=30+x => x=43-30=13

2. Nice proof, Peter!

3. Geometric Solution of the problem 830 by Michael Tsourakakis from Greece
See the image:http://img820.imageshack.us/img820/2807/geogebra.png
solution
CM is perpendicular of AD. Because CA=CD , CM is mediator of the AD.
AB and MC meet of G. So triangle AGD is isosceles .So , angle GDA=73
E is point of the GM with AED is equilateral triangle.
Angle GDE= angle EAG =73-60=13, therefore , angle EDB=73-34-13=26
Additional, angle DAB=73 and angle BDA=34 ,so , angle ABD=73,therefore DA=DB ,so ,DB=DE and triangle BDE is isosceles.Therefore,2 angle DBE=1800-260=1540 so, angle DBE= angle DEB=77.But angle CBD=30, therefore, angle EBC=47
But, angle BEC=77- angle CED=77-30=47.Therefore , triangle BCE is isosceles and CB=CE .But DB=DE .So , DC is mediator of the BE ,therefore, DE is bisector of the angle BDE=26.So,x = angle CDB=13

4. Draw CF and DE perpendicular to AD and BCrespectively, Then easily FD equal to DE and CFD and CED Tr.s are congruent.Hence < CDE = x + 34. so 2x + 34=60. So x=13.

Sumith Peiris
Moratuwa
Sri Lanka

5. Problem 830
Form the equilateral triangle ADE ( the point C is inside the triangle ADE).So EC is bisector