By barycentric coordinates, as $I_a=\left[-\frac{a}{b+c-a},\frac{b}{b+c-a};\frac{c}{b+c-a}\right]$ we get that $$S_{(I_aBC)}=\frac{a}{b+c-a}S_{(ABC)}=\frac{a}{b+c-a}\sqrt{\frac{a+b+c}{2}\cdot \frac{b+c-a}{2}\cdot \frac{a+b-c}{2}\cdot \frac{a+c-b}{2}}=MW\cdot \frac{a+b+c}{2}$$iff $$MW=\frac{a}{2}\sqrt{\frac{a^2-(b-c)^2}{(b+c)^2-a^2}}$$ $M=\left[0,\frac{1}{2},\frac{1}{2}\right]$ and $W=\left[-\frac{a^2}{(b+c)^2-a^2}, \frac{b(b+c)}{(b+c)^2-a^2}, \frac{c(b+c)}{(b+c)^2-a^2}\right]$ Then the displacement vector WM has the following coordinates: $$x=-\frac{a^2}{(b+c)^2-a^2}, y=\frac{a^2+b^2-c^2}{2[(b+c)^2-a^2]}, z=\frac{a^2+c^2-b^2}{2[(b+c)^2-a^2]}$$ $$MW^2=-a^2yz-b^2xz-c^2xy=\frac{a^2}{4}\left(\frac{-a^4-b^4-c^4+2a^2b^2+2a^2c^2+2b^2c^2}{[(b+c)^2-a^2]^2}\right)=\frac{a^2}{4}\cdot \frac{a^2-(b-c)^2}{(b+c)^2-a^2}$$ Hence we are done

Solved with L567 Let $K$ and $L$ be the the foot of perpendicular from $I_a$ to $AC$ and $BC$. Note that $\triangle KAI_a \sim \triangle MBW$. Thus, we have that $$MW \cdot p = MW \cdot AK = BM \cdot I_aK = \frac{1}{2}BC \cdot I_aL$$QED.