In the diagram below, $AE=AF$ and $BM=MC$. Show that $\frac{EP}{FP}= \frac{AC}{AB}$ 
I have tried using Menelaus but I am not sure that it is correct. My thought was to use menelaus in different triangles in order to cancel and then get the ratio I needed but I couldnt get it.
 A: *

*Area of triangle ABM is the same as AMC (same perpendicular to BC and BM=MC) 

*Area of triangle PBM is the same as PMC (same reasoning)

*From 1 and 2, area of ABP is the same as area pf APC

*Draw perpendicular from P to AB (say Q) and from P to AC (R). Then PQAB=PRAC.

*The angle AEF is the same as AFE

*From 4 and 5 EPQ and PRF are proportional (all angles are the same)

*From 6 PR/PE=PQ/PF. We can rewrite it as EP/FP=PR/PQ

*From 3 and 4 PRAB=PQAC so PR/PQ=AC/AB

*& and 8 give you your solution

A: From the Menelaus Theorem on $\triangle BEK$ and line $A-P-M$ we have:
$$\frac{BA}{EA} \times \frac{EP}{PK} \times \frac{KM}{BM} = 1$$
Similarly from the Menelaus Theorem on $\triangle CFK$ and line $A-P-M$ we have:
$$\frac{CA}{FA} \times \frac{FP}{KP} \times \frac{KM}{CM} = 1$$
Equating the two equations we have:
$$\frac{BA}{EA} \times \frac{EP}{PK} \times \frac{KM}{BM} = \frac{CA}{FA} \times \frac{FP}{KP} \times \frac{KM}{CM}$$
$$\frac{BA}{CA} = \frac{FP}{EP}$$
Hence the proof.
A: 
Noting that $\triangle APE=\frac{1}{2}EA\times PA\times\sin(\angle EAP)$ and $\triangle APE=\frac{1}{2}FA\times PA\times\sin(\angle PAF)$:
\begin{align}
\frac{EP}{PF}&=\frac{\triangle APE}{\triangle APF}=\frac{\sin(\angle EAP)}{\sin(\angle PAF)}=\frac{\sin(\angle BAM)}{\sin(\angle MAC)}\\
&=\frac{\sin(\angle BAM)}{BM}\Big/\frac{\sin(\angle MAC)}{MC}\\
&=\frac{\sin(\angle AMB)}{AB}\Big/\frac{\sin(\angle AMC)}{AC}\\
&=\frac{AC}{AB}.
\end{align}
The second line uses $BM=MC$, the third the Law of Cosines, and the last $\sin(x)=\sin(\pi-x)$.
A: Here is proof without Menelaus' theorem or trigonometry. It uses only geometric constructions and the intercept theorem (Thales' theorem). 


*

*Let $b$ be the line that passes through point $B$ and is parallel to line $EF$. Let $b$ intersect line $AC$ at point $B'$. Then $BB'=b$ is parallel to $EF$ and so by the intercept theorem $AB = AB'$ because $AE = AF$.

*Let $P'$ be the intersection point of lines $AM$ and $BB'$. Since $BB'$ is parallel to $EF$, by the intercept theorem 
$$\frac{BP'}{B'P'}=\frac{EP}{FP}$$    



*Let $l$ be the line through point $B$ and parallel to $AM$. Let $l$ intersect line $AC$ at point $C'$. Then $BC' = l$ is parallel to line $AM$.

*Since $MB = MC$ and $BC' \, \| \, AM$, by the intercept theorem it follows that 
$$1 = \frac{MC}{MB} = \frac{AC}{AC'}$$ and thus $AC' = AC$.

*Finally, since $BC' \,\|\, AP'$ because $P'$ lies on $AM$, by the intercept theorem and the conclusions in points 1. and 4. it follows that
$$\frac{BP'}{B'P'} = \frac{AC'}{AB'} = \frac{AC}{AB}$$
Hence, by the relation in point 2.  $$\frac{EP}{FP} = \frac{BP'}{B'P'} = \frac{AC}{AB}$$
