## Sunday, May 18, 2008

### Elearn Geometry Problem 10 See Problem 10 details at:
www.gogeometry.com/problem/problem010.htm

Triangle, Angles, Congruence. Level: High School, SAT Prep, College geometry

1. bae deok rak(bdr@krea.com)October 3, 2010 at 9:01 PM

Let O be the circumcenter of the triangle ABC.
Let E be the intection of the extension of BO and CA, where O is an circumcent of ABC.
Let F be the point on the extension of CO with triangle ABF is an equilateral triangle. Then we have ang(FBO)=10 deg=ang(ECO),
ang(FOB)=20 deg=ang(EOC) and OB=OC(circumradius) and so two triangles OFB and OEC are ongruence.
Hence we get CE=FB=AB, that is, E=D and hence
x= ang(DBC)=10 degree.

2. Greetings

3. For the sake those who possibly couldn’t follow Newzad’s excellent solution, I re-write it here: Draw a line CE below AC such that /_ACE=40 and CB=CE so that Tr. ECB is equilateral which makes BE=BC. Now Tr. ABE is SAS congruent with Tr. DCB for AB=CD,BE =BC and /_ABE=20=/_DCB and therefore /_x =/_AEB. Further, AC=BC=CE and thus Tr. ACE is isosceles with an apex angle=40 and thus /_AEC = x+60 = 70 which makes x=10.

4. I think the solution can further be simplified if we
“draw the equilateral triangle BCE on the same side as A”.

CBA forms an 20-80-80 isosceles triangle with CB = CA.

angle(ABE) = 80 – 60 = 20 = angle(DCB)
AB = DV and EB = BC
Therefore, [BAE] is SAS-congruent to [CDB].
Thus, angle AEB = x

CB = CA = CE implies [CAE] is an 40-(x + 60)-(x+ 60) isosceles triangle.
In [CAE], 40 + (x + 60) + (x+ 60) = 180
Therefore, x = 10

Mike from Canotta

1. 5. There is also a solution if we draw the equilateral triangle BCE on the opposite side as A. Then by SAS triangle CED is congruent to triangle ABC, so AC=BC=BE=EC=ED, so BED is isoceles and since angle BED=40, angle BDE=70, so angle x=10.

6. 7. I hope that the followıng solutıon ıs not among those posted on varıous lınks shown above:

Construct equilateral triangle ABE, inside the triangle ABC; CE is angle bisector of <ACB and easily triangles BCE and CBD are congruent, (s.a.s.), so <CBD= <BCE= 10 degs.

Best regards,
Stan Fulger

1. it does not seem so evident
regards
Daniel

8. Align triangle BCD so CD overlays AB and call the new vertex E. This leaves an inner 60 degree angle at EAC . AC = BC since angle BAC = ABC = 80. So the new edge EA = BC since we just copied the triangle and transitively EA = BC = AC. Since EA = AC, and EAC is 60 triangle EAC is equilateral.
Then angle ECB is 40 and the triangle its in is also isosceles. So Angle BEC = angle CBE which by angle chasing means 60 + x = 80 -x or x = 10.

9. 10. 11. 