Hypothesis: Every triangle is "n-dissectable" (can be divided into n isosceles triangles) if $n \ge 4$.
Every triangle can be dissected into two right triangles (how?): by dissecting it along the altitude to its longest side. Each of these can be dissected into two isosceles triangles (why?): by cutting along the median from the right angle to the hypotenuse in each, giving a dissection of the original triangle into 4 isosceles triangles.
Now use induction on $n$ to prove the case that any triangle can be divided into 5 isosceles triangles.
To prove it draw a line from any of the vertex to opposite side to get triangles so that one of the triangles is isosceles. Consider the remaining triangle as the starting triangle and based on what we know about being 4-dissectable, that any triangle can be divided to 4 isosceles triangles, we get 4 isosceles triangles from this triangle, (the second of the two from the of original triangle. This gives us that the first isosceles triangle, plus the 4 sub-isosceles triangles combine to make any triangle dissectable (divided) into a total of 5 isosceles triangles.
An equilateral triangle can be dealt with separately: Cut off a small equilateral triangle from a corner (base parallel to to larger base), and since an isosceles trapezoid always has a circumcircle, one can obtain 4 isosceles triangles from within the trapezoind = total 5. [Alternatively, if you count the original equilateral triangle as a separate isosceles triangle, the proceed as in the case for $n = 4$ to obtain $4$ interior isosceles triangles, you have, in total, 5 isosceles triangles.]
The attachment can clarify matters further.
See Dissecting a triangle into 4 or more isosceles triangles
The above link goes into great detail and proves what you need to prove, and provides pictures that portray exactly what's going on, which might help, in terms of any language-translation barriers.