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Restate the problem: Let $a,b,c$ be side lengths of triangle $ABC, b<c$. Let $D$ be a point on $BC$ such that $AD$ is the interior angle bisector of $\angle A$. (1)Find a necessary and sufficient condition, in terms of $\angle A,\angle B,\angle C$, such that there exist points $E$ and $F$ satisfying the following condtions: $BE=CF, \angle BDE=\angle CDF$, where $E,F$ lie on $AB,AC$(not the vertices) respectively. (2)Find $BE$ in terms of $a,b,c$, assuming there exist such points $E,F$.

My solution: Let $x=\widehat{BZE}=\widehat{CZF}$ Applying Sines Law into triangles $CYZ$ and $CXB$: $\frac{CF}{sinx}=\frac{CZ}{sin(C+x)}$ $\frac{BE}{sinx}=\frac{BZ}{sin(B+x)}$ But $CF=BE$ so we conclude that: $\frac{CZ}{BZ}=\frac{sin(C+x)}{sin(B+x)}$ But $AZ$ is the angle bisector of $\widehat{BAC}$ so: $\frac{CZ}{BZ}=\frac{AC}{AB}=\frac{sinB}{sinC}$ So $sinB.sin(B+x)=sinC.sin(C+x)$ $\Rightarrow cos(2B+x)=cos(2C+x)$ Note that $2B+x<2.90+90=270, 2C+x<2(C+x)<360$.So we must have $2B+x+2C+x=360$, which leads to $x=A$. Since $F$ lies on $AC$ we must have $x\leq\widehat{AZC}=\frac{A}{2}+B$ Hence we must have $\frac{A}{2}\leq B$, which is also the sufficient condition.

For b/: $\frac{BZ}{CZ}=\frac{AB}{AC}$ $\Rightarrow\frac{BZ}{BC}=\frac{AB}{AB+AC}$ Note that $\frac{BE}{sinA}=\frac{BZ}{sinC}$ and $\frac{BC}{AB}=\frac{sinA}{sinC}$, we conclude that: $BX=\frac{BC}{AB}.BZ=\frac{a^2}{b+c}$ Q.E.D