- 17 Feb '08 03:24Let 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. - 17 Feb '08 13:55

Doesnt seem to be enough info?*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.

Is that ANY point D on BC or midpoint?

And E could be anywhere depending on where the 20degree angle on AD is taken from. ... or is EDA 20? - 17 Feb '08 15:48let me clarify a bit -

Going clockwise, mark points A, B, C such that AB = BC = CA. (i.e , the triangle is equilateral).

1- take A: The side opposed to A is BC. Draw a perpendicular line from A to BC. This line will intersect BC at point D - Note that because the line is perpendicular to BC => BD = DC.

2- Now, take the segment AD. Draw a 20 degrees segment (having AD as reference) that starts at D and intersects AB at a point E, i.e. the angle ADE is 20 degrees.

3- Link E to C.

4- Determine angle DEC

I hope this helps - 17 Feb '08 16:19

I understand the question now! Whether that helps or not .....*Originally posted by smaia***let me clarify a bit -**

Going clockwise, mark points A, B, C such that AB = BC = CA. (i.e , the triangle is equilateral).

1- take A: The side opposed to A is BC. Draw a perpendicular line from A to BC. This line will intersect BC at point D - Note that because the line is perpendicular to BC => BD = DC.

2- Now, take the segment AD. Draw a 20 degrees segme ...[text shortened]... .e. the angle ADE is 20 degrees.

3- Link E to C.

4- Determine angle DEC

I hope this helps - 17 Feb '08 18:41 / 2 editsDEC = 32.55º

I gave the equilateral triangle sides with 9 units.

AD = 7.8

AE = 3.48

EC = 7.86

ACE = 22.55º

ECD = 37.45º

First I worked out AD. With AD you can solve AE. With AE you can work out EC. With EC you can work out angle ACE.

Angle ECD = 37.45º (60 - 22.5). Angle EDC (110) plus angle ECD (37.45) = 147.45º. The sum of the angles of a triangle = 180º -147.45º = 32.55º.

Is this correct? - 17 Feb '08 19:46

Nice try, but the answer is not 32.55*Originally posted by Green Paladin***DEC = 32.55º**

I gave the equilateral triangle sides with 9 units.

AD = 7.8

AE = 3.48

EC = 7.86

ACE = 22.55º

ECD = 37.45º

First I worked out AD. With AD you can solve AE. With AE you can work out EC. With EC you can work out angle ACE.

Angle ECD = 37.45º (60 - 22.5). Angle EDC (110) plus angle ECD (37.45) = 147.45º. The sum of the angles of a triangle = 180º -147.45º = 32.55º.

Is this correct?

If you want to follow your approach, you need to set AB = x where x is any real number.

Good luck. - 17 Feb '08 20:52

In two weeks I will post the solution here.*Originally posted by Green Paladin***Ok, I've checked my numbers. Final attempt (with no rounding off) - 32.725229402º**

You will be amazed how beautiful and simple the solution is.

No need to use calculators or perform complex computations.

It can be resolved mentally if you discover a few hidden things.