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^{\circ}

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 = \frac{a}{2}

and

OZA = 90^{\circ}

Hence $AOZ = 90 - \frac{a}{2}$

Now consider triangle XOZ. THis triangle is isosceles, so $OXZ = XZO = \frac{180 - (180 - a)}{2} = \frac{a}{2}$
By similar arguments $OXY = OYZ = \frac{b}{2}$ and $OYZ = OZY = \frac{c}{2}$ Hence the new angles of the triangle are $ZXY = \frac{a}{2} + \frac{b}{2}$

$XYZ = \frac{b}{2} + \frac{c}{2}$

$YZX = \frac{c}{2} + \frac{a}{2}$

$a + b + c = 180$

Hence $\frac{a + b}{2} = 90 - \frac{c}{2}$

It follows that $ZXY = 90 - \frac{c}{2}$ .

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: $90 - \frac{a}{2}$

triangle 3: $90 - \frac{90 - \frac{a}{2}}{2}$ = $90 - \frac{90}{2} + \frac{a}{4}$

triangle 4: $90 - \frac{90 - \frac{90}{2} + \frac{a}{4}}{2} = \frac{3}{4} . 90 - \frac{a}{8}$

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.