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

Post your solutions or ideas in the comments.

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. Similar problem... http://www.youtube.com/watch?v=4S4RJGk6Uxs

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

1. please explain why CE=BD ?
it does not seem so evident
regards
Daniel

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.

11. 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°

12. 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.

13. 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