1. ## Find X

upd: here's a pic...

ΔABC is isosceles.
Find X WITHOUT using trig or algebra, by constructing segments || to those given and such and deriving angles from there, knowing that vertically opposite angles are congruent, etc...

I already solved this prolbem using sine and cosine laws, but that was not acceptable ... Spend days on this one, even tried solving backwards having X be known (from solving it the other way), but still nothing...

i don't want the solution, but if you could just point me in the right direction, get me started, i'd appreciate it!

2. Originally Posted by Solarmew
upd: here's a pic...

ΔABC is isosceles.
Find X WITHOUT using trig or algebra, by constructing segments || to those given and such and deriving angles from there, knowing that vertically opposite angles are congruent, etc...

I already solved this prolbem using sine and cosine laws, but that was not acceptable ... Spend days on this one, even tried solving backwards having X be known (from solving it the other way), but still nothing...

i don't want the solution, but if you could just point me in the right direction, get me started, i'd appreciate it!
You should be able to figure it out if you realise that the sum of the angles in a triangle is $180^\circ$ and a straight line makes $180^\circ$.

To get you started, since $\triangle ABC$ is isosceles, then $\angle ABC$ and $\angle ACB$ are equal.

3. haha, i know that ... it's just that i already scribbled up ten pages drawing parallel lines here and there and yet i can't seem to find any angles that are helpful...
at this point my papers look like a tokyo subway map... only much much more confusing X.X

kinda like this only a lot more lines and numbers...

here's the proof i came up with... seems legit to me >.> ... i'll use "=~" as "similar to" and "/_" as "angle" for this occasion X.X ...

if:

ΔABC =~ ΔCFG =~ΔAEH
then
/_BGD = /_DBE = /_DEO = 30°

5. any comments? does that look right? or at least as a starting point?

6. Originally Posted by Solarmew
here's the proof i came up with... seems legit to me >.> ... i'll use "=~" as "similar to" and "/_" as "angle" for this occasion X.X ...

if:

ΔABC =~ ΔCFG =~ΔAEH
then
/_BGD = /_DBE = /_DEO = 30°
That is an ingenious construction, and it has the big advantage of giving the correct answer (which I can only get by using trigonometry).

But I don't see how the construction works. Starting from ABC, you construct two similar triangles, a small one AEH and a large one FGC. It's clear how to get the first of these, you just choose H on the line AC so that AH = AE. But how do you construct the large triangle? It looks as though you construct FG to be the line through B parallel to AE. Then F is the point where this line meets AC (extended), and presumably G is chosen so that FG = FC. But then how do you know that D, E and G are collinear? And why is the angle BDG equal to 30º? Either I'm misunderstanding something, or more explanation is needed!

7. yeah it definitely looks pretty bare ... i didn't think this was going to be it, i was just hooping someone who's done this before could tell me if im on the right track so that i dun waste my time trying to make smth work that's just not ment to X.X ...
i guess i'll try anyways... see if i can throw some more angles in there, maybe it'll magically make more sense XD

as far as D, E and G be colinnear...
instead of drawing a bigger similar triangle outside of ABC and a smaller one inside, we could also draw two smaller similar triangles, one inside ABC and the second inside the first one following the same procedure, this way we would be sure that all the lines are straight as we wouldn't have to extent any, but instead use the ones already given...
so technically its the same thing