Solving Isosceles Triangle ABC: Finding Angle CDE

[Ursole]

An isosceles triangle ABC has a point D on AB and a point E on AC.

Given:
Angle BAC = 20 degrees
Angle EBC = 50 degrees
Angle DCB = 60 degrees.
AB = AC

Find angle CDE.

 

[rattis]

you could do this by scale drawing, but i can't be bothered to. I drew out some rough shetches and got a few angles but not the one you wanted.

 

[Reddhawk]

I'd say its 70 degrees

 

[davilla]

This problem is evil.

Please tell me that the answer is at least a multiple of 10 degrees. Using sine law and some of the symmetry in the figure, I got it down to (sin x)/(sin x+20) = (sin 80)/(sin 40) or something like that, but I don't know how to solve such a complex equation without using a graphing calculator.

 

[Ursole]

Davilla, your equation is correct.

Besides the trignometrical solution, there are as many as eight more elegant consructive proofs.

And, if one is really a masochist, it is possible to attempt a general solution in terms of the three given angles.

 

[Gokul43201]

I'd say its 70 degrees

I think not...

 

[Gokul43201]

This problem is evil.

Please tell me that the answer is at least a multiple of 10 degrees. Using sine law and some of the symmetry in the figure, I got it down to (sin x)/(sin x+20) = (sin 80)/(sin 40) or something like that, but I don't know how to solve such a complex equation without using a graphing calculator.

For 0<x<90, sin(x+20) > sin(x). So this can't give you a solution in 0<x<90. And clearly (a reasonable drawing will show) x is in this range. So, it's probably the "something like that" that's right.

For all my talking, I'd better at least, start showing some results...

 

[Gokul43201]

Davilla, your equation is correct.

Besides the trignometrical solution, there are as many as eight more elegant consructive proofs.

And, if one is really a masochist, it is possible to attempt a general solution in terms of the three given angles.

Since you suggest "constructive proofs" I think you could start by finding some point P1 on one of AB or AC such that triangle BCP1 is isosceles. On top of this construct more such isosceles triangles. If you're lucky, I think D or E will coincide with some Pn, from which you can calculate angles geometrically.

I "think" this may true. I tried it and found it to be true...but when I rechecked I found it was not...I'm not going to check again now...perhaps later.

 

[Ursole]

I got it down to (sin x)/(sin x+20) = (sin 80)/(sin 40) or something like that,

Sorry, I meant 'something like that' is correct.

 

[davilla]

I said "something like that" because I was writing from memory. Anyways I labeled a different angle as x.

Besides the trignometrical solution, there are as many as eight more elegant consructive proofs.

If that's true then I'm convinced there is a right angle in the figure, resulting in even more symmetry. Working backwards from my presumption there are at least seven formulas that, if could be derived from the statements in the problem, would prove that solution. The simplest is to show that 2 * DE = BE. But putting trig aside, the miraculous hidden line, if it exists, is elusive!

Perhaps I'm looking for something that's too easy. After all there are three or four statements needed to define the problem. If no two of them combine in a simplified form, then we're basically dealing with a system of several equations. Either I'm too lazy to churn them or this problem has exposed the limits of my atrophied cognitive capacity!

 

[Ursole]

Working backwards from my presumption there are at least seven formulas that...

No, I meant that there are eight different ways to arrive at the correct answer without using trig. formulae.

 

[Pfft]

Thirty degrees. Sneaky!

 

[Ursole]

Thirty degrees. Sneaky!

You have posted an answer, whereas I was hoping for a solution.

 

[ceptimus]

Call the bottom of the main triangle side A and the line running from the bottom left corner to theta side B.

Then, by the sin law on the bottom-right triangle,
A/sin40 = B/sin80
A/B = sin40/sin80

Now, notice that the bottom-left triangle is icosceles (80-50-50), so
A/sin(theta) = B/sin(160-theta)
A/B = sin(theta)/sin(160-theta)

Nw we can combine these to get sin(theta)/sin(160-theta) = sin(40)/sin(80)

theta=30.

 

[Pfft]

Oh!

The answer is straight forward, I'll get it to you this time tomorrow.

Pfft

 

[Pfft]

Given:
An isosceles triangle ABC has a point D on AB and a point E on AC.

/_BAC=20 degrees
/_EBC=50 degrees
/_DCB=60 degrees
AB=AC

Find angle CDE

By definition ABC =ACB so ABC and ACB each equal 80 degrees.
ACB-CBE=DBE=30 degrees
By the same logic ACB-BCD=DCE=20 degrees
180-CBE-BCD-DCE=BEC=50 degrees

Note that lines DC and BE form a cross inside of the ABC triangle. We will call that cross point X.
180-BCE-BCD=70=BXC and therefore CXE is also equal to 70 degrees.

Hypothesis: If line BC is extended from C by a distance equal to CD and that distant point were called Y, and a second line is drawn from E to Y and called EY, then a triangle CDY will be formed which may be another isosceles triangle. If is true, then angle ACY=120 degrees and so is CDE+CYE=60. CDE would then equal 30 degrees.

Proof: If CDY were indeed an isosceles triangle, then line segment EY would be a straight line continuation of line DE. Since angle EXD=70 degrees and we wish to prove that CDE=30 degrees, it must be that DEX=80 degrees.
180-BEC-DEX=50=CEY. and CDY+ECY+EYC=180 removing any doubt that the assumption that CDY is an isosceles triangle.

Therefore it must be true that CDE=30 degrees.

 

[Ursole]

Check out http://mathcircle.berkeley.edu/BMC4/Handouts/geoprob.pdf for some interesting historical background to the puzzle, plus EIGHT different solutions.