Question

Two circles intersect each other in points X and Y.Secants through X and Y intersects one of the circles in A and D and the other in B and C respectively. A and B are on the opposite sides of line DC.Show that line AD is parallel to line BC.

Answer 1

$$\begin{gathered} \angle \mathrm{ADY}+\angle \mathrm{AXY}=180°(\mathrm{ADYX}\mathrm{is}\mathrm{a}\mathrm{cyclic}\mathrm{quadrilateral}) \\ \angle \mathrm{BCY}+\angle \mathrm{BXY}=180°(\mathrm{BXYC}\mathrm{is}\mathrm{a}\mathrm{cyclic}\mathrm{quadrilateral}) \\ \mathrm{By}\mathrm{adding}\mathrm{both}\mathrm{we}\mathrm{get}, \\ \angle \mathrm{ADY}+\angle \mathrm{BCY}+\angle \mathrm{AXY}+\angle \mathrm{BXY}=360° \\ \Rightarrow \angle \mathrm{ADY}+\angle \mathrm{BCY}+\angle \mathrm{CAB}+\angle \mathrm{BAD}=360° \\ \Rightarrow \angle \mathrm{ADY}+\angle \mathrm{BCY}+180°=360°\left(\angle \mathrm{AXY}+\angle \mathrm{BXY}=180°,\mathrm{linnear}\mathrm{pair}\right) \\ \Rightarrow \angle \mathrm{ADY}+\angle \mathrm{BCY}=180° \end{gathered}$$

Since. the angles on the same side of a transversal line are supplementary
Hence,$\mathrm{AD}\parallel \mathrm{BC}$