David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, $A,\text{ }B,\text{ }C$, which can each be inscribed in a circle with radius $1$. Let $\varphi_A$ denote the measure of the acute angle made by the diagonals of quadrilateral $A$, and define $\varphi_B$ and $\varphi_C$ similarly. Suppose that $\sin\varphi_A=\frac{2}{3}$, $\sin\varphi_B=\frac{3}{5}$, and $\sin\varphi_C=\frac{6}{7}$. All three quadrilaterals have the same area $K$, which can be written in the form $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Firstly, fora concyclic quadrilateral, the following relation holds where $R$ is the circumradius. $$K=\frac{\sqrt{(ab+bc)(ac+bd)(bd+ac)}}{4R}$$ Now, let the two diagonals be $m$ and $n$, respectively. Then, $$K=\frac{1}{2}mn\sin\varphi_A=\frac{1}{2}mn\sin\varphi_B=\frac{1}{2}\sin\varphi_C$$ Applying Ptolemy theorem and replace $mn$ in the above relation leads to $$K=\frac{1}{2}(ab+cd)\frac{2}{3}=\frac{1}{2}(ac+bd)\frac{3}{5}=\frac{1}{2}(ad+bc)\frac{6}{7}$$ This means that $$(ab+cd)=3K, (ac+bd)=\frac{10}{3}K, (ad+bc)=\frac{7}{3}K$$ Setting these to the first relation and solve $K$ gives $$K=\frac{24}{35}\implies m+n=\boxed{59}$$