i think everyone is reading too much into it, pretty sure angle DEC=50 degrees

Originally posted by smaia
Nice try, but sorry, this is not the correct answer. You don't need to develop such amazingly complicated calculations to solve this problem.
I assure you: all you need is
1- the sum of the internal angles of a triangle is 180
2- The properties of similar triangles
3- the opposite angles of two intersecting lines are equal and other axioms of euclidian geometry

Good luck.

Originally posted by greent
i think everyone is reading too much into it, pretty sure angle DEC=50 degrees

Originally posted by Aetherael
don't fault me for saying so... but i'm starting to doubt your proposed solution, unless i'm missing something very important in the diagram. what similar triangles are you talking about?

Originally posted by mtthw
I'm having similar thoughts. Unless there's still some misunderstanding about the problem.

This is what I think it should look like:

http://aycu01.webshots.com/image/45880/2000590380518787909_rs.jpg

Any comments?

Originally posted by mtthw
I'm having similar thoughts. Unless there's still some misunderstanding about the problem.

This is what I think it should look like:

http://aycu01.webshots.com/image/45880/2000590380518787909_rs.jpg

Any comments?

Originally posted by greent
great picture, thats how mine looks on paper, i took your picture and added the angles i think should be in there,
http://www.sendspace.com/file/g1cd8m

Originally posted by mtthw
Given that it's drawn pretty much to scale, you don't find it strange that your 20 degree angle at ECD is bigger than your 40 degree angle at ECA?

(Oh, and that file as a GIF is about 200 times smaller! 🙂)

Originally posted by mtthw
Again, irrelevant. It's about similifying your expressions.

I got (first attempt - will check it later):

DEC = arctan[sin(50)sin(70)/(sin(50)cos(70) + sin(60))]

No calculators in sight. Clearly not the 'simple' method you're looking for, though.

Originally posted by smaia
The question is: what is the angle?
Assume you have a multiple choice test:
A- 10
B- 20
C- 30
D- 40
E- 50

Without using a calculator what is the correct answer?

smaia, come on, give us the answer, its driving me nuts.*EDIT* please message me the answer

Originally posted by greent
smaia, come on, give us the answer, its driving me nuts.*EDIT* please message me the answer

How can there be an elegant solution with an angle of ~32.54 degrees?

http://s117.photobucket.com/albums/o51/Jirakon/?action=view&current=Triangle.png

(1) a/sinz = b/sin50
(2) a/sin(70-z) = b/sin60

(2)/(1) => sinz/sin(70-z) = sin50/sin60

z = 32.54

Originally posted by smaia
Let ABC be a triangle where AB = BC = AC.
From A, draw a line that intersects BC at a point D.
From the AD segment, draw a 20 degrees line that intersects AB at a point E. Determine the angle DEC.
This is not an easy problem.
Hint: Try to see invisible things.
Good luck. And consider yourself a genius if you resolve it in less than 30 minutes. Don't feel bad about yourself if you don't resolve it at all.

the resolution (the hard way, without triangle axioms):
- D is a middle point of BC
- angle EDC is 110º
- imagine a point F, middle point of AB (the line FD will be parallel to AC)
- angle DCF is 30º and angle FCA is 30º too
- line EA is 2 times bigger then line EF (because angle FDE is 10º and angle EDA is 20º😉
- so, angle FCE is 10º (because FCA is 30º and is bisected by E in proportion 2:1)
- Therefore, DCE is 40º
- Therefore, DEC is 180-110-40=30

But there's an axiom for all of this... I made 10x more calculations then necessary just to show the hard way