Find the perimeter of triangle $ABC$. In triangle $ABC$, point $M$ is the midpoint of side $BC$. Suppose that the point
$D$ and $E$ on the $CA$ side so $CD = DE$ with $D$ between $C$ and $E$.
Suppose also points $F$ and $G$ on the side $AB$ so that $AM$, $BD$, and $CF$ meet at one point. Likewise with $AM$, $BE$, and $CG$. If, $AF + BM + CD = 17$ and $AG + BM + CE = 23$. Find the perimeter of triangle $ABC$.

With Ceva's Theorem obtained
$-$ $EG || DF || CB$
$-$ $\Delta ABC$ ~ $\Delta AEG$ ~ $\Delta ADF$
$GF=BF$
Let $AG = ax$, $AE = bx$, $GF = ay$, $ED = by$, $BM = z$
$ax + ay + z + by = 17$
$ax + z + 2by = 23$,
We get $y (b-a) = 6$
Perimeter $\Delta ABC$ = $(\frac{x}{y}-1)+40$
But i can find the value of $x/y$. if someone can help that great!

Let $AG/AB = AE/AC = r$.  Then because $F$ is the midpoint of $GB$, we have $$\frac{AF}{AB} = \frac{AG + (AB - AG)/2}{AB} = \frac{1+r}{2}.$$  Similarly, $$\frac{CE}{AC} = \frac{AC - AE}{AC} = 1 - r, \quad \frac{DC}{AC} = 1 - \frac{1+r}{2} = \frac{1-r}{2}.$$  Therefore, $$17 = AF + BM + CD = \frac{1+r}{2} c + \frac{a}{2} + \frac{1-r}{2} b, \\ 23 = AG + BM + CE = r c + \frac{a}{2} + (1-r) b,$$ where $a, b, c$ are the respective sides of $\triangle ABC$.  Then multiplying the first equation by $2(1-2r)/(1-r)$ and the second by $2r/(1-r)$, we get $$a+b+c = 22 + \frac{12}{1-r}.$$  This suggests that the answer depends on $r$.
To prove that the triangle is not uniquely determined, we can illustrate two distinct examples.
Example 1. Choose $r = 1/4$; we get $a+b+c = 38$.  Then choose $a = b = 18$, $c = 2$.  Consequently $AG = 1/2$, $BM = 9$, $CE = 27/2$, $AF = 5/4$, and $CD = 27/4$.
Example 2.  Choose $r = 1/13$; we get $a+b+c = 35$.  Then choose $a = 14$, $b = 17$, $c = 4$.  Consequently $AG = 4/13$, $BM = 7$, $CE = 204/13$, $AF = 28/13$, and $CD = 102/13$.
It is worth noting that we must have $0 < r < 5/11$, otherwise the triangle inequality is not met.
