Is it possible to divide a isosceles triangle We have an isosceles triangle and we divide it by two sections going out of one of three corners, hence we get three new triangles. Is it possible to make (puzzle) an isosceles triangle out of every two of three triangles that we got after dividing the first one? (In case you would ask me what I tried, I begged my teacher for an answer. Yes, it is possible, but the proof is still needed) Thank you in advance!
 A: I'm not sure if I get the Question right, but it should not, in general,
be possible (I don't know how to make a picture here, so I will try to
explain it as best as possible).
Take an icosceles triangle, call it $\Delta(A,B,C)$ where $A,B \quad \text{and} \quad C$ are the names of the corners. 
Now, let's start with Point C, and draw 2 lines from there which will intersect 
$\overline{AB}$ at some Points $C_1$ and $C_2$.
Take the corresponding triangles $\Delta(A,C_1,C)$, $\Delta(C_1,C_2,C)$ and
$\Delta(C_2,B,C)$.
Now, If I understand you correctly, you want to know if it is possible to ''glue'' any two of these two triangles together and get an icosceles triangle.
In general, this should not be possible, take for example $\Delta(A,C_1,C)$ and 
$\Delta(C_1,C_2,C)$, we know that the lenght $L(\overline{CC_1})$ of $\Delta(A,C_1,C)$ equals $L(\overline{CC_1})$ of $\Delta(C_1,C_2,C)$ but nothing
about the remaining side-lenghts. In particular, you can (try it at home!) divide it in such a way, that $L(\overline{AC_1}) \neq L(\overline{C_1C_2})$ 
and $L(\overline{CA}) \neq L(\overline{CC_2})$. Thus, the only possible way to glue the two of these together would be to glue at the $\overline{CC_1}$ line!
But this results in the triangle $\Delta(AC_2C)$ hence, if you split $\Delta(A,B,C)$ in such a way that $L(\overline{AC_2}) \neq L(\overline{AC})$ there is no possible way to acieve your goal (the latter will always be the case since your triangle was icosceles so any nontrivial choice of sections will give this result).
A: It is possible (at least for some isosceles triangles).

Take an isoceles triangle $ABC$ with $AB = AC$ and $36^\circ < \angle BAC < 45^\circ$.


*

*Let the perpendicular bisector of $AC$ intersect $AB$ at $D$. 

*Draw a circle at $C$ through $D$, it will intersect $AB$ at $D$ and $E$. 

*Cut $\triangle ABC$  along $CD$ and $CE$, we get two new isosceles  $\triangle ADC$ and $\triangle CDE$
(by construction, $AD = CD = CE$ ).

