Question

In the given figure B and E are points on the line segment AC and DF respectively. Show that AD || CF.

Answer 1

ABED and BCFE are cyclic quadrilaterals
$\angle CBE$ is external angle to ABED
$\Rightarrow \angle CBE=\angle ADE$
But $\angle CBE+\angle CFB={180}^{\circ }$
$\angle ADE+\angle CEB={180}^{\circ }$
Similarity
$\angle BAD+\angle BCF={180}^{\circ }$
$\therefore$ AC is transversal between AD and CF
and $\angle BAD$ and $\angle BCF$ are internal angles and their sum is ${180}^{\circ }$
$+\angle ADE$ and $\angle CFE$ are internal angles and their sum is ${180}^{\circ }$
$\Rightarrow$ AD is parallel to CF.