I don't think this warrants an insight, so here are the two methods. My epiphany was realizing that once the first piece is chosen, the third apex of all valid triangles must lie on an ellipse.
Method 1
Let a be the first piece with an upper limit of half the perimeter p (p = (a+b+c)/2). The sum of the other two pieces is constant and equal to 2p - a. This means that the third vertex of the triangle must lie on an ellipse, the separation betwen the two foci of which is the base of the triangle a. The distance r from one of the foci to the ellipse is given in polar coordinates as
##r(\phi)=\frac{C}{1+\epsilon \cos \phi}##
where ##\epsilon## is the eccentricity. We can find constant C by setting ##\phi## = 0 in which case ##r(\phi)## has minimum and maximum values,
##r_{max}=\frac{C}{1-\epsilon}##; ##r_{min}=\frac{C}{1+\epsilon}##
Note that ##r_{max}-r_{min}=a## from the definition of these quantities for an ellipse.
But also when ##\phi = 0##, ##2r_{min}+a=2p-a \rightarrow r_{min}=p-a## which gives ##C=(1+\epsilon)(p-a)## in which case
##r(\phi)=\frac{(1+\epsilon)(p-a)}{1+\epsilon \cos \phi}##
The eccentricity is the ratio of the base (distance between foci) to the sum of the two other sides,
##\epsilon=\frac{a}{2p-a}## so that, finally,
$$r(\phi)=\frac{2p(p-a)}{(2p-a)+a \cos \phi}$$
The probability that the first piece be between ##a## and ##a+da## is uniform and given by
##dP_a=\frac{1}{p}da## subject to the constraint ##0<a<p##.
The probability that the second piece be between ##r## and ##r+dr## is also uniform and given by
##dP_r=\frac{1}{r_{max}-r_{min}}dr=\frac{1}{a}dr## subject to the constraint ##r_{min}<r<r_{max}##.
The latter probability in terms of ##\phi## is
##dP_{\phi}=\frac{1}{a} \frac{\partial r}{\partial \phi}d\phi=\frac{2p(p-a)\sin \phi}{[(2p-a)+a \cos \phi]^2}d \phi##
The area of the triangle is
##A(r,\phi)=\frac{1}{2}\left| \vec{a} \times \vec{r}(\phi) \right |=\frac{ap(p-a)\sin \phi}{(2p-a)+a \cos \phi}##
The expectation value for the area is
$$\left< A \right>=\int{A(a,\phi)dP_a~dP_{\phi}}=2p\int_0^p{a(p-a)^2 da} \int_0^\pi{\frac{\sin^2 \phi}{[(2p-a)+a \cos \phi]^3} d \phi}$$Integration yields
$$\left< A \right>=\frac{\pi p^2}{30}.$$
Method 2
Heron's formula
##A=\sqrt{p(p-a)(p-b)(p-c)}~;~p=(a+b+c)/2##
can be transformed to
##A=\frac{1}{2}\sqrt{p(p-a)(a^2-\eta^2)}~;~\eta \equiv b-c##
where ##a## is the first broken piece. As before, the probability that the first piece be between ##a## and ##a+da## is uniform and given by
##dP_a=\frac{1}{p}da## subject to the constraint ##0<a<p##.
The remaining piece can be broken into two more pieces of which the larger piece is labeled ##b## and the smaller piece is labeled ##c##. We define the difference between the pieces ##\eta \equiv b-c##. The probability that ##\eta## be between ##\eta## and ##\eta+d\eta## is uniform and given by ##dP_{\eta}=\frac{1}{a}d\eta## subject to the constraint ##0<\eta<a##
The expectation value for the area is
$$\left< A \right>=\int{A(a,\eta)dP_a~dP_{\eta}}=\frac{1}{2p} \int_0^p{\frac{1}{a}da} \int_0^a{\sqrt{p(p-a)(a^2-\eta^2)}d\eta}$$
Integration yields$$\left< A \right>=\frac{\pi p^2}{30}.$$