lengths of sides in golden ratio isosceles triangles The figure below shows three different isosceles triangles. every triangle is either 36-36-108 or 36-72-72. The base of the outermost triangle has length $\phi$. 

Find the lengths of both lines AT and MT.
Can someone please help me figure this out? What I'm thinking so far is that $1+\phi=\phi^2$ has roots which are the golden ratio.
I found this image online (on wikipedia) and i think it is the solution but i don't know how to solve it or actually get these values


As shown on Wikipedia's article about golden ratio (letters in the blockquote indicate points on the picture provided by Wikipedia):  

If angle BCX = α, then XCA = α because of the bisection, and CAB = α because of the similar triangles; ABC = 2α from the original isosceles symmetry, and BXC = 2α by similarity. The angles in a triangle add up to 180°, so 5α = 180, giving α = 36°. So the angles of the golden triangle are thus 36°-72°-72°. The angles of the remaining obtuse isosceles triangle AXC (sometimes called the golden gnomon) are 36°-36°-108°.

From the given above (in the question) $AH=TH= \phi$ and $m(\angle AHT)=36°$, So by using Law of cosines:
$${AT}^2={AH}^2+{TH}^2-2(AH)(TH)(\cos {\angle AHT}) \\  {AT}^2={\phi}^2+{\phi}^2-2(\phi)(\phi)(\cos {36°}) \\ {AT}^2=2{\phi}^2-2\phi^2 \cdot \cos {36°} \\ {AT}^2=2 \left( {\frac {3+ \sqrt 5}{2}} \right) -2 \left( {\frac {3+ \sqrt 5}{2}} \right) (\cos {36°})=1 \\ AT=\sqrt 1 =1$$
Similarly, 
$${AC}^2=2 {\phi}^2-2{\phi}^2(\cos 108°)={\phi}^4 \\ AC = {\phi}^2$$
Now you know it, law of cosines is the trick in such questions.  
I hope my answer helps you !  
Another solution:-
You can use trigonometrical functions to get the height of any of the two triangles and then use Pythagoras theorem to find other sides (in each triangle).
