First, there's the special case of an isosceles right triangle, which can be split into two such similar triangles by the bisector of the right angle.

A more general case (that does not include the special case above) is one in which the line from one of the vertices to the opposite side creates two isosceles triangles where: 

In one of the two triangles the two equal sides do not include the line segment that constitutes the split.

In the other of the two triangles the segment that performs the split is equal to the segment cut off on the opposite side that is along the same line as one of the sides of the first triangle.

Call the base angles of the first formed isosceles triangle b, and the base angles of the second isosceles triangle a.

The apex angle of the first isosceles triangle is 180° - 2*b on the basis of that isosceles triangle. But as part of the large triangle (the one being split), it's equal to 180 - a - (a+b). Equating the two indicates b = 2a. The split divides the angle at which it divides the large triangle, which is a+b, is therefore 3*a, while the other angle shared with the second isosceles triangle is a.

So the relation sought is that one of the angles is three times another. Take as an example a triangle with angles 10°, 30° and 140°. Split the 30°-angle so a 10° angle pairs with the original 10° angle to form an isosceles triangle. The remaining 20° fraction of the original angle pairs with a 20° newly formed angle supplementary to the 160° apex angle of the 10-10-160 triangle.