Quadrilateral $ABCD$ with chord of inscribed circle meeting $AC$ at $P$. Find ratio $AP$ : $PC$

Question

In a quadrilateral $ABCD$, there is an inscribed circle centered at $O$. Let $F,N,E,M$ be the points on the circle that touch the quadrilateral, such that $F$ is on $AB$, $N$ is on $BC$, and so on. It is known that $AF=5$ and $EC=3$. Let $P$ be the intersection of $AC$ and $MN$. Find the ratio $AP:PC$.

I know that $AM=AF=5$ and $CN=CE=3.$ The answer is equivalent to the ratio of the areas $[ADP]:[DPC]$. I cannot continue on from this point. Would anyone please help?

Answer 1

COMMENT:

As can be seen ratio $\frac{AP}{PC}$ depends on the positions of C related to A. So more constrains must be included, for example " A, O(center of circle) and C are colinear, otherwise there can be infinite solutions.

Answer 2

Construct line from C parallel to the chord MN, meeting the extension of AD at X. Then, XMNC is isosceles, implying XM = CN = CE. Note that the triangles AMP and AXC are similar, leading to

$$\frac {AP}{PC} =\frac {AM}{MX} = \frac {AF}{CE} = \frac 53 $$

Answer 3

Here's a ridiculously high-powered solution, because why not :)

By Brianchon's theorem on the "hexagon" $ABCEDM$, we get that $AE, BD, CM$ concur. So by Ceva's theorem, $$\frac{AP}{PC}\cdot\frac{CE}{ED}\cdot\frac{DM}{MA}=1\implies\frac{AP}{PC}=\frac{MA}{CE}=\frac{5}{3}.$$