## 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.

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

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. 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

7. 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. Solution:

10. Here's my solution.