Meg offers this solution:

The tangent of a circle is at right-angles to the radius of the circle. That is, if you join the centre point of the circle to a point where the circle meets the outer triangle, it makes an angle of 90 ° with the side of the triangle.

triangle with inscribed circle and radius
The bisector of the angles of triangle ABC will all pass through the centre of the circle.
triangle with inscribed circle and bisectors
From this we know that

OAZ = a
2

and OZA = 90 °

Hence
AOZ = 90- a
2

Now consider triangle XOZ. THis triangle is isosceles, so
OXZ = XZO = 180-(180-a)
2
 = a
2


By similar arguments
OXY = OYZ = b
2

and
OYZ = OZY = c
2

Hence the new angles of the triangle are
ZXY = a
2
 + b
2


XYZ = b
2
 + c
2


YZX = c
2
 + a
2

a+b+c=180

Hence
a+b
2
 = 90 - c
2

It follows that
ZXY = 90 - c
2

.

A similar argument can be followed for XYZ and YZX. If you continue drawing triangles within circles, the angles will decrease as shown here:

Triangle 1: a

triangle 2:
90 - a
2

triangle 3:
90-
90- a
2
2

=
90 - 90
2
 + a
4

triangle 4:
90-
90- 90
2
 + a
4
2
 = 3
4
 .90 - a
8

When you continue this iteration, you can demonstrate that the a term becomes less and less significant, and the sum of the rest of the terms tends to 60 degrees. Hence the triangle tends to an equilateral triangle.