We have $BD = \frac{BM}{2cos(MBC)}= \frac{BM}{\frac{BM^{2}+a^{2}-\frac{b^{2}}{4}}{a.BM}}$ $= \frac{a.BM^{2}}{BM^{2}+a^{2}-\frac{b^{2}}{4}}$ So, $BD.DC = BD(a-BD) = \frac{a^{2}BM^{2}(a^{2}-\frac{b^{2}}{4})}{(BM^{2}+a^{2}-\frac{b^{2}}{4})^{2}}= S$ i) We know that $BM^{2}= \frac{2(a^{2}+c^{2})-b^{2}}{4}$, so, $(BM^{2}+a^{2}-\frac{b^{2}}{4})^{2}= (\frac{a^{2}+c^{2}-b^{2}}{2}+a^{2})^{2}= (\frac{c^{2}+bc+a^{2}}{2})^{2}$ $= (\frac{2c^{2}+2bc+b^{2}+bc}{4})^{2}= \frac{(b+c)^{2}(2c+b)^{2}}{16}$ ii) We have $a^{2}.BM^{2}(a^{2}-\frac{b^{2}}{4}) = \frac{a^{2}(2a^{2}+2c^{2}-b^{2})}{4}.(\frac{b^{2}+bc}{2}-\frac{b^{2}}{4})$ $= \frac{a^{2}(2c^{2}+bc)}{4}.\frac{b^{2}+2bc}{4}= \frac{a^{2}bc(2c+b)^{2}}{16}$ By i) and ii), we have that $S = \frac{a^{2}bc}{(b+c)^{2}}= \frac{b^{2}c}{2(b+c)}$!

Remark. Using the above simple lemma this problem becomes very easily and "painted nastily" !

let $P$ be a point on line $AC$ and at side of segment $AC$ such that $AP=AB$ $2BC^{2}-AC^{2}=AB.AC \longleftrightarrow CM.CP=BC^{2}$ so circumcircle of $\triangle PBM$ is tangent to $BC$ at $B$ and cause $DM=BD$ , $DM$ is tangent to the circle ,too. we have $\angle ABP=\angle MPB=\angle DBM=\angle DMB$ hence $\angle MDC=\angle BDM+\angle BMD=\angle A$ so quadrilateral $MDBA$ is cyclic,hence $\angle DBM=\angle DMB=\angle CAD=\angle BAD$ so $AD$ is anglebisector of $\angle A$ therfore $AD$ is perpendicular to $BM$ so $c=AM=\frac{b}{2}$ we have: $CM.AC=CD.BC \rightarrow \frac{b^{2}}{2}=\frac{ab}{b+c}.a \Longrightarrow a^{2}=3c^{2}=\frac{3b^{2}}{4}$$\star$ so $BD.DC=\frac{AC^{2}.AB}{2(AB+AC)}$

My solution is a little bit similar as Amir.S 's: Let $B'$ be the piont on the line $AC$ and $B'A=BA$ and $B',A,C$ lie in order. Let $A'$ be the point on the line $BC$ with the properth $A'B=CB$ and $A'B,C$ lie in order. then $2BC^{2}-AC^{2}=AB*AC-\to A,B,A',B$lie on a circle and $\triangle CAB \sim \triangle CA'B'$ and we easy to check that $\triangle CB'B\sim \triangle CBM,\triangle BMD \sim \triangle B'BA \triangle MDC\sim \triangle BAC$ in fact $A-\to A',B-\to B',M\to M'=B, D--\to D'=A$ then we have that $\frac{BD*DC}{B'A*AC}=(\frac{BC}{B'C})^{2}=\frac{AC^{2}AB}{2(AB+AC)}$

Amir.S wrote: therfore $ AD$ is perpendicular to $ BM$ Why is this?

radio wrote: Amir.S wrote: therfore $ AD$ is perpendicular to $ BM$ Why is this? yes,thanks. that was wrong ,but I didn't use it in rest of solution,Proving that $ AMDB$ is cyclic is enough to complete the solution.

How, Amir.S? So you won't get $ c=\frac{b}{2}$...

radio wrote: How, Amir.S? So you won't get $ c = \frac {b}{2}$... That's not nessecery also. as i see know the problem is more easier than I thuaght,The relation $ CB^2 = CM.CP$ and the fact that D lies on angle bisector is enough to solve the problem. this relation is equal to $ a^2 = \frac {b(b + c)}{2}\Longleftrightarrow \frac {a^2bc}{(b + c)^2} = \frac {b^2c}{2(b + c)}\Longleftrightarrow BD.DC = \frac {AC^2.AB}{2(AB + AC)}$cause:$ BD = \frac {ac}{b + c}\ ,DC = \frac {ab}{b + c}$ thanks again -Amir.S

A nice problem Consider point $ K$ on ray $ CA$, in which $ AK=AB$, Simply we can show that $ CM.CK=CB^2$ It means $ \triangle CBM$ and $ \triangle CKB$ are similar to each other,and: $ \angle CBM=\angle AKB \Longleftrightarrow ABDM$is cyclic. According to as inversion lemma we have: $ A'B'=\frac{k.AB}{OA.OB}$ (which $ k$, as radius and $ O$,as the center of inversion,and$ A',B'$ the inverse points of $ A,B$respectively) then: $ AB=\frac{\frac{AC^2}{2}.BD}{CD.\frac{AC}{2}}=\frac{AC.BD}{CD}=\frac{2CM.BD}{CD}$ and: $ AC=2CM$ according to these equalities,lets find out the value of $ CD.BD$: $ CD.BD\Longleftrightarrow \frac{CD.CB.BD}{CB}\Longleftrightarrow \frac{CM.CA.BD}{CB}$ $ \Longleftrightarrow \frac{2CM^2.BD}{\frac{BC}{CD}.CD}$ $ \Longleftrightarrow \frac{(4CM^2)(2CM.BD)}{2.2CM(\frac{BC}{BD}).CD}$ $ \Longleftrightarrow \frac{(4CM^2)(2CM.BD)}{2.2CM(\frac{BD}{CD}+1).CD}$ $ \Longleftrightarrow \frac{(4CM^2)(2CM.BD)}{2(\frac{2CM.BD}{CD}+2CM).CD}$ $ \Longleftrightarrow \frac{(4CM^2)(2CM.BD)}{2(AB+AC).CD}$ $ \Longleftrightarrow \frac{AC^2.AB}{2(AB+AC)}$ and we proved that: $ CD.BD= \frac{AC^2.AB}{2(AB+AC)}$ hmmm,but I really don’t like it!