Answer. $\frac{AF}{AD}=\frac{2}{5}$ Claim. $KD=KA$ Proof. Angle chasing gives $\angle CBK=\angle BKA=\angle KBA$. Hence, $KA=AB=\frac{BC}{2}=\frac{AD}{2}$ or $KA=KD$ Now that $AK=KD$ or $\frac{DK}{AK}=1$. Applying Menelaus' theorem on $\triangle ADP$ with transversal $M,L,F$ gives $$\frac{AF}{FD}\cdot\frac{DM}{PM}\cdot\frac{PL}{LA}=1$$By BPT $$\frac{DM}{PM}=\frac{KB}{LB}=\frac{AK+AD}{AD}=\frac{3}{2}$$and $$\frac{PL}{LA}=\frac{DK}{KA}=1$$Thus, $\frac{AF}{FD}=\frac{3}{2}$ or $\frac{AF}{AD}=\frac{2}{5}$

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Set the origin of the coordinate at $B$ with $ \angle BAD=\theta$ and let lines $BC$ and $DM$ intersect at $K_1$.It is obvious that $K$ and $K_1$ are the midpoints of $AD$ and $BC$ respectively. Let $AB=a$. With this, we know: $A(-a\cos\theta,a\sin\theta),B(0,0),C(2a,0)$ and $D(2a-a\cos\theta,a\sin\theta)$. Additionally, $K(a-a\cos\theta,a\sin\theta)$ , $K_1(a,0)$. $M$ is the reflection of $A$ with respect to $B$. So, $M(a\cos\theta,-a\sin\theta)$ The position of $L$ can be found as it is the intersection of segments $BK$ and $AC$. Equation of $BK \Rightarrow \frac{x}{a-a\cos\theta}=\frac{y}{a\sin\theta} \cdots (1)$ Equation of $AC \Rightarrow \frac{x+a\cos\theta}{a\cos\theta+2}=\frac{a\sin\theta-y}{a\sin\theta} \cdots (2)$ Combining $(1)$ and $(2)$ and further simplifications will give the coordinates of $L(\frac{2a-2a\cos\theta}{3},\frac{2a\sin\theta}{3})$ $F$ is the intersection of $ML$ and $AD$. Equation of $ML \Rightarrow \frac{x-a\cos\theta}{a\cos\theta-\frac{2a-2a\cos\theta}{3}}=\frac{y+a\sin\theta}{-a\sin\theta-\frac{2a\sin\theta}{3}}$ Simplifying, $5x\sin\theta=y(2-5\cos\theta)+2a\sin\theta \cdots (3)$ Equation of $AD$ is simply $y=a\sin\theta \cdots (4)$ Solving $(3)$ and $(4)$ gives the coordinates for $F(\frac{4a-5a\cos\theta}{5},a\sin\theta)$ So, $AF=a\cos\theta+\frac{4a-5a\cos\theta}{5}=\frac{4a}{5}$ And $ \boxed {\frac{AF}{AD}=\frac{2}{5}}$