From Menelaus' theorem $\frac{AN}{NC}\frac{CB}{BD}\frac{DM}{MA}=1$=>$\frac{CN}{AN}=2$=> $AN=\frac{AC}{3}=\frac{b}{3}$=>$CN=\frac{2}{3}b$=> $BN^2*b=BC^2*AN+AB^2*CN-AC*AN*CN=\frac{1}{9}(3a^2b+6c^2b-2b^3)$=> $BN^2=\frac{1}{9}(3a^2+6c^2-2b^2)$ Again from Menelaus' theorem we have $\frac{BM}{MN}\frac{NA}{AC}\frac{CD}{DB}=1$ =>$\frac{BM}{MN}=3$=> $\frac{BM}{MN}=\frac{BC^2}{BN^2}$<=> $\frac{BC^2}{BN^2}=3$<=>$b^2=3c^2$. Also the condition "$AB$ is tangent to the circuncircle to the triangle $NBC$"<=>$AB^2=AC*AN$<=>$c^2=\frac{1}{3}b^2$<=> $b^2=3c^2$.

The problem is much easier than you think, Tiks... We will work with directed segments. The Menelaos theorem for triangle BNC and the collinear points A, M, D on its sidelines NC, BN, BC yields $\frac{BM}{MN}\cdot\frac{NA}{AC}\cdot\frac{CD}{DB}=-1$. Since $\frac{NA}{AC}=-\frac{NA}{CA}$ and $\frac{CD}{DB}=1$ (the latter because D is the midpoint of the segment BC), this becomes $\frac{BM}{MN}\cdot\left(-\frac{NA}{CA}\right)\cdot 1=-1$, what simplifies to $\frac{BM}{MN}=\frac{CA}{NA}$. On the other hand, let the tangent to the circumcircle of triangle NBC at the point B meet the line CA at a point K. Then, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=6557 post #5 Lemma 2, applied to the triangle NBC, we have $\frac{CK}{NK}=\frac{BC^2}{BN^2}$. Now, if the line AB is tangent to the circumcircle of triangle NBC, then this line AB is the tangent to the circumcircle of triangle NBC at the point B; hence, the line AB meets the line CA at the point K; in other words, A = K. This yields $\frac{CA}{NA}=\frac{CK}{NK}$. Hence, from $\frac{BM}{MN}=\frac{CA}{NA}$ and $\frac{CK}{NK}=\frac{BC^2}{BN^2}$, it follows that $\frac{BM}{MN}=\frac{BC^2}{BN^2}$. Conversely, if $\frac{BM}{MN}=\frac{BC^2}{BN^2}$, then, since $\frac{BM}{MN}=\frac{CA}{NA}$ and $\frac{CK}{NK}=\frac{BC^2}{BN^2}$, we obtain $\frac{CA}{NA}=\frac{CK}{NK}$. Thus, the points A and K divide the segment CN in the same (signed) ratio; therefore, they must coincide. Hence, the line AB coincides with the line KB. Since the line KB is the tangent to the circumcircle of triangle NBC at the point B, it thus follows that the line AB is this tangent to the circumcircle of triangle NBC. Altogether, we have shown that the line AB is tangent to the circumcircle of triangle NBC if and only if $\frac{BM}{MN}=\frac{BC^2}{BN^2}$. Thus, the problem is solved. We have not used the condition that the point M is the midpoint of the segment AD; our solution shows that the point M could have been any arbitrary point on the line AD instead. darij

Darij I don't think that this problem is hard ,I just had written all ideas which was in my minde and the problem was solve.Anyway thank you for another solution,but this problem can be named as an exersisa in geometry ,so we shouldn't much discusse it .

Tiks and Darij, I invite you to an amusement ! Suppose that I don't know the Menelaus' theorem. A preliminary study. Define the middlepoint $P$ of the segment $[CN]\ (PN=PC)\ .$ $1.\ \triangle BNC\ : \ DB=DC\ ,\ PN=PC\Longrightarrow DP\parallel BN\equiv MN\ ,\ \boxed {BN=2\cdot DP}\ .$ $2.\ \triangle ADP\ : \ MA=MD\ ,\ MN\parallel DP\Longrightarrow NA=NP\ ,\ \boxed {DP=2\cdot MN}\ .$ $3.\ BN=2\cdot DP=4\cdot MN\Longrightarrow BM+MN=4\cdot MN\Longrightarrow \boxed {\frac{MB}{MN}=3}\ .$ $4.\ NA=NP=PC\Longrightarrow \frac{AC}{AN}=3\Longrightarrow \boxed {\frac{MB}{MN}=\frac{AC}{AN}}=3\ .$ The proper proof. The line $AB$ is tangent to the circumcircle of the triangle $BNC$ iff $\triangle ABC\sim \triangle ANB$ $\Longleftrightarrow$ $\frac{AB}{AN}=\frac{BC}{NB}=\frac{CA}{BA}$ $\Longleftrightarrow$ $\left(\frac{BC}{NB}\right)^2=\frac{AB}{AN}\cdot \frac{CA}{BA}$ $\Longleftrightarrow$ $\left(\frac{BC}{BN}\right)^2=\frac{AC}{AN}$ $\Longleftrightarrow$ $\left(\frac{BC}{BN}\right)^2=\frac{MB}{MN}\ .$ Remark. Darij, the your observation - the point $M$ is freely on the median $[AD]$ - is remarkably ! Why find you fault with Tiks ? He made no one mistake. Maybe you wish a solution as my above proof !

B is tangent to the circuncircle ABC <=> <ABN=<ACB=a and <=> AB^2 = AN*AC (3) <ABN=<ACB=a <=> sin(a)/sin(<BNC)= BN/BC and sin(a)/sin(<ANB)=AN/AB, but when sin(<BNC)=sin(<ANB) <=> BN/BC=AN/AB <=> BC/BN=AB/AN (1) <=> (BC/BN)^2=(AB/AN)^2 (2) BN=2BK and BC=2DC, where K is the midpoint of BN, BC/BN=AB/AN <=> BK/DC=AN/AB <=> AB*BK=AN*DC (4) if divide (3) to (4).. <=> AB/BK=AC/DC <=> ABK is similar to ADC <=> <BAK=<DAC <=> AM is simedian of ABN <=> BM/MN=(AB/AN)^2 and by (2) <=> BM/MN=(BC/BN)^2

mehhhhhhhhhhh First note that $ABC\sim ANB$, because of the tangency. Then, $BC/BN=AC/AB=AB/AN$ $(i)$. Note that also $BM/MN=(BC/BN)^2 \leftrightarrow BM/MN=(AB/AN)^2$. But law of sines on $ABM,ANM$ gives us: $BM/BN=AB.sin \angle BAM/AN.sin\angle NAM.= (AB/AN)^2$. So we just need $AB/BN=sin \angle BAM/sin \angle NAM$. But $\angle BAM=\angle BAD, \angle NAM=\angle CAD$, and again by law of sines on $ABD,ACD$ we get that: $sin \angle BAD/sin \angle CAD=AC/AB$. So we just need $AB/BN=AC/AB$, which is true by $(i)$.