Could someone explain me how to demonstrate the following problem ?
(I had to translate it so sorry for any grammatical mistakes as english isn't my first language)
From a point C outside a circumference draw a tangent CA and a secant that intersect the circumference in B and D with CB larger than CD. From a point E on the secant with CE lesser than CD draw the parallel to AB that intersects the tangent in F. Demonstrate that $\angleCAD = \angleDEF$ and that the quadrilateral ADEF can be inscribed.
Thank you in advance
$\angle CAD=\angle DBA$ (angle in alternate segment)
$\angle CEF=\angle DBA$ (corresponding angles of parallel lines)
So, $\angle CAD=\angle CEF$.
$ADEF$ is a cyclic quadrilateral is a consequence of $\angle CAD=\angle CEF$. (exterior angle is equal to interior opposite angle)