The opposite angles of a cyclic quadrilateral are supplementary

$$\begin{array}{rl}\text{m}\mathrm{\angle }\text{a}+\text{m}\mathrm{\angle }\text{c}& ={180}^{\circ }\\ \\ \text{m}\mathrm{\angle }\text{b}+\text{m}\mathrm{\angle }\text{d}& ={180}^{\circ }\\ \end{array}$$

If T is the point of intersection of the two diagonals, then $\text{AT}\times \text{TC}=\text{BT}\times \text{TD}$