## 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. Solution:

11. Here's my solution.

12. This solution is much clearer when illustrated, so grab a pen and a piece of paper.

Here are some starting info:
> Angle ABC = 180° - 20° - 80° = 80°
> Triangle ABC is isosceles as angles ABC and BAC are equal.
> Line AC = line BC because triangle ABC is isosceles.

Add point E to the figure and connect it to points A, B and C, such that triangle BCE is equilateral. Do note that triangle BCE should overlap triangle ABC. Line AC = line BC = line BE = line CE.

Firstly, focus on triangle ACE.
> Triangle ACE is isosceles because line AC = line CE.
> Angle ACE = 60° - 20° = 40°
> Angle CAE = angle CEA = (180° - 40°)/2 = 70°

Next, focus on triangle ABE.
> Angle ABE = 80° - 60° = 20°
> Angle AEB = 70° - 60° = 10°

Lastly, notice that triangles ABE and BCD are congruent by SAS:
> Line AB = line CD
> Angle ABE = angle BCD = 20°
> Line BE = line BC

Angle AEB = angle CBD = x because triangles ABE and BCD are congruent.

x = 20°

1. 13. Angle ABC = 180 - 20 - 80 = 80
Add point E such that CDE is congruent to ABC. Angle BCE = 80 - 20 = 60. Also note that angle CED is 20 due to congruence.
We have BC congruent to CE and a 60 degree angle between, so connecting EB forms an equilateral triangle BCE. Angle BED = 60 - 20 = 40.
Now notice that triangle BED is isosceles, meaning angle EBD is (180 - 40) / 2 = 70. Therefore x is 70 - 60 = 10.

14. Consider triangle BCD
sin20/BD=sinx/CD
BD/CD=sin20/sinx-------(1)

Consider triangle ABD
<ABD=20+x
sin80/BD=sin(20+x)/AB
BD/AB=sin80/sin(20+x)----------(2)

By equating (1) & (2)
sin20/sinx=sin80/sin(20+x)
sin20/sin80=sinx/sin(20+x)
2sin10cos10/cos10=sinx/sin(20+x)
2sin10sin(20+x)=sinx
cos(10+x)-cos(30+x)=sinx
cos10cosx-sin10sinx-cos30cosx+sin30sinx=sinx
cosx(cos10-cos30)=sinx(1+sin10-sin30)
tanx=(cos10-cos30)/(1+sin10-sin30)
x=10