Let the triangle be $\triangle A B C$, with $A C$ as the hypotenuse, and let $D, E, F, G$ be on sides $A B, B C, A C$, $A C$, respectively, such that they all lie on the circle. We have $A G=1, G F=24$, and $F C=3$. By power of a point, we have $$\begin{aligned} & A D=\sqrt{A G \cdot A F}=\sqrt{1(1+24)}=5 \\ & C E=\sqrt{C F \cdot C G}=\sqrt{3(3+24)}=9 \end{aligned}$$ Now, let $B D=B E=x$. By the Pythagorean Theorem, we get that $$\begin{aligned} (x+5)^{2}+(x+9)^{2} & =28^{2} \\ (x+5)^{2}+(x+9)^{2}-((x+9)-(x+5))^{2} & =28^{2}-4^{2} \\ 2(x+5)(x+9) & =768 \\ (x+5)(x+9) & =384 \end{aligned}$$ The area of $\triangle A B C$ is $\frac{1}{2}(x+5)(x+9)=\frac{1}{2} \cdot 384=192$.