Prove that DEC ||| BEA and FAB ||| FCA given that ABCD is a cyclic quadrilateral and FA is tangent to the circle at A.

$$\angle E = \angle E$$

This is a common angle to both triangles $\triangle DEC$ and $\triangle BEA$.

ABCD is a cyclic quadrilateral. The sum of opposite angles in a cyclic quadrilateral is $180^\circ$.

$$\angle BAD + \angle BCD = 180^\circ$$

Angles on a straight line sum to $180^\circ$.

$$\angle DCE + \angle BCD = 180^\circ$$

From these two equations, we can conclude:

$$\angle DCE = \angle BAD$$

Since two angles of $\triangle DEC$ are equal to two angles of $\triangle BEA$, the triangles are similar by the Angle-Angle (AA) similarity criterion.

$$\boxed{\triangle DEC ||| \triangle BEA}$$

$$\angle F = \angle F$$

This is a common angle to both triangles $\triangle FAB$ and $\triangle FCA$.

FA is a tangent to the circle at A. The angle between the tangent FA and the chord AB ($\angle FAB$) is equal to the angle in the alternate segment subtended by the chord AB ($\angle ACB$ or $\angle FCA$).

$$\angle FAB = \angle FCA$$

Since two angles of $\triangle FAB$ are equal to two angles of $\triangle FCA$, the triangles are similar by the Angle-Angle (AA) similarity criterion.

$$\boxed{\triangle FAB ||| \triangle FCA}$$

Since $\triangle FAB ||| \triangle FCA$, the ratio of their corresponding sides is equal:

$$\frac{FA}{FC} = \frac{AB}{CA} = \frac{FB}{FA}$$

From the first two parts of the ratio:

$$\frac{FA}{FC} = \frac{AB}{CA}$$

Cross-multiplying gives:

$$FA \cdot CA = FC \cdot AB$$

This completes the proof.

$$\boxed{FA \cdot CA = FC \cdot AB}$$

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$$\angle J = \angle J$$

This is a common angle to both triangles $\triangle JRS$ and $\triangle JTK$.

KRST is a cyclic quadrilateral. The sum of opposite angles in a cyclic quadrilateral is $180^\circ$.

$$\angle KRS + \angle KTS = 180^\circ$$

Angles on a straight line sum to $180^\circ$.

$$\angle JTK + \angle KTS = 180^\circ$$

From these two equations, we can conclude:

$$\angle JTK = \angle KRS$$

Since two angles of $\triangle JRS$ are equal to two angles of $\triangle JTK$, the triangles are similar by the Angle-Angle (AA) similarity criterion.

$$\boxed{\triangle JRS ||| \triangle JTK}$$

Since $\triangle JRS ||| \triangle JTK$, the ratio of their corresponding sides is equal:

$$\frac{JR}{JT} = \frac{RS}{TK} = \frac{JS}{JK}$$

From the first two parts of the ratio:

$$\frac{JR}{JT} = \frac{RS}{TK}$$

Cross-multiplying gives:

$$JR \cdot TK = JT \cdot RS$$

We are given that $JT = RS$. Substitute this into the equation:

$$JR \cdot TK = RS \cdot RS$$ $$JR \cdot TK = RS^2$$

This completes the proof.

$$\boxed{JR \cdot TK = RS^2}$$