Y by Rounak_iitr
Let ABCD be a cyclic quadrilateral, which is AB=7 and BC=6. Let E be a point on segment CD so that BE=9. Line BE and AD intersect at F. Suppose that A, D, and F lie in order. If AF=11 and DF:DE=7:6, find the length of segment CD.
Stay ahead of learning milestones! Enroll in a class over the summer!
As $ABCD$ is cyclic $\angle EDF = \angle ABC$, both supplementary to the same angle $\angle CDA$. Hence $ABC \sim EDF$ and $\angle AFB = \angle DFE= \angle EBC=\angle FBC$, so $AD \parallel AF \parallel BC$. Since every cyclic trapezoid is isosceles, it follows that $\boxed{\overline{CD}=7}$.
As $ABCD$ is cyclic $\angle EDF = \angle ABC$, both supplementary to the same angle $\angle CDA$. Hence $ABC \sim EDF$ and $\angle AFB = \angle DFE= \angle EBC=\angle FBC$, so $AD \parallel AF \parallel BC$. Since every cyclic trapezoid is isosceles, it follows that $\boxed{\overline{CD}=7}$.
Something appears to not have loaded correctly.