27
21
24
30
18

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$\angle{AFB}=\angle{ADB}=90^{\circ}$ , quadrilateral $ABDF$ is cyclic. It follows that $\angle{ADE}=\angle{ABF}$ .

$\angle{AFB}=\angle{AED}=90\implies$ triangles $ABF \text{ and} ADE$ are similar.

It follows that $AF=(13)(\frac{4}{5}), BF=(13)(\frac{3}{5})$ .

By Ptolemy, we have $13DF+(5)(13)(\frac{4}{5})=(12)(13)(\frac{3}{5})$ . Divide the whole equation by $13$ to obtain

$DF=\frac{16}{5}\implies{16+5= \boxed{21}}.$

Ptolemy Theoremgives a relationship between the side lengths and the diagonals of a cyclic quadrilateral; it is the equality case of Ptolemy's Inequality. Ptolemy's Theorem frequently shows up as an intermediate step in problems involving inscribed figures. The statement is as follows:Given a cyclic quadrilateral $ABCD$ with side lengths ${a},{b},{c},{d}$ and diagonals ${e},{f}$ :

$ac+bd=ef.$