Who know the solution of this problem? :

A nice and difficult (for who doesn't know the lower lemma) metrical problem ! anonymous1173 wrote: $ D$ is a point on the edge $ BC$ of triangle $ ABC$ such that $ AD = \frac {BD^2}{AB + AD} = \frac {CD^2}{AC + AD}$. Denote the point $ E$ is a point such that $ D\in [AE]$ and $ CD = \frac {DE^2}{CD + CE}$. Prove that $ AE = AB + AC$. Proof. The relations from hypothesis suggest me to apply the well-known property : Quote: Lemma. Let $ ABC$ be a triangle. Denote the point $ D\in BC$ for which $ AD\perp BC$ . Suppose $ D\in (BC)$ . Then $ b^2 = c(a + c)\Longleftrightarrow B = 2C\Longleftrightarrow DC = DB + BA$ . Proof. $ \boxed {\ b^2 = c(a + c)\ }\Longleftrightarrow b^2 - c^2 = ac\Longleftrightarrow$ $ DC^2 - DB^2 = ac\Longleftrightarrow$ $ \boxed {\ DC = DB + BA\ }\Longleftrightarrow$ $ b\cos C = c + c\cdot \cos B$ $ \Longleftrightarrow$ $ \sin B\cos C = \sin C(1 + \cos B)$ $ \Longleftrightarrow$ $ \tan\frac B2 = \tan C$ $ \Longleftrightarrow$ $ \boxed {\ B = 2C\ }$ . Denote $ \|\begin{array}{c} m(\widehat {ABC}) = x \\ \\ m(\widehat {ACB}) = y \\ \\ m(\widehat {CED}) = z\end{array}$ . Apply the upper lemma to the triangles : $ \|\begin{array}{cccc} \triangle\ ABD\ : & BD^2 = AD\cdot (AB + AD) & \implies & m(\widehat {BAD}) = 2\cdot\widehat {ABD}) = 2x \\ \\ \triangle\ ACD\ : & CD^2 = AD\cdot (AC + AD) & \implies & m(\widehat {CAD}) = 2\cdot m(\widehat {ACD}) = 2y \\ \\ \triangle\ CDE\ : & DE^2 = CD\cdot (CE + CD) & \implies & m(\widehat {DCE}) = 2\cdot m(\widehat {CED}) = 2z\end{array}$ . Observe that $ A + B + C = 180^{\circ}$ $ \Longleftrightarrow$ $ (2x + 2y) + x + y = 180^{\circ}$ $ \Longleftrightarrow$ $ \boxed {\ x + y = 60^{\circ}\ }$ , $ m(\widehat {ABD}) + m(\widehat {BAD}) = m(\widehat {CED}) + m(\widehat {DCE})$ $ \Longleftrightarrow$ $ x + 2x = z + 2z$ $ \Longleftrightarrow$ $ \boxed {\ z = x\ }$ and $ ABEC$ is cyclically. Thus, $ AE = AB + AC$ $ \Longleftrightarrow$ $ \sin \widehat {ACE} = \sin\widehat {ACB} + \sin\widehat {ABC}$ $ \Longleftrightarrow$ $ \sin (2x + y) = \sin y + \sin x$ $ \Longleftrightarrow$ $ \sin (2x + y) - \sin y = \sin x$ $ \Longleftrightarrow$ $ 2\sin x\cos (x + y) = \sin x$ $ \Longleftrightarrow$ $ \cos (x + y) = \frac 12$ $ \Longleftrightarrow$ $ x + y = 60^{\circ}$ , what is truly.

We have $ AD=\frac{BD^2}{AB+AD}\Rightarrow AB\cdot AD = BD^2 - AD^2 $. Let the circle with center $D$ and radius $DA$ cut $BA$ at $X$. The power of $B$ is $BD^2 - DA^2 = BA \cdot BX$. So $BX = AD = XD \Rightarrow \angle BAD = 2\cdot \angle ABD = 2\alpha$. Similarly, $\angle DAC = 2\cdot \angle ACD = 2\beta$. We have $\alpha + \beta + 2\alpha + 2\beta = 180^\circ \Rightarrow \alpha + \beta = 60^\circ$. Again similarly, and since $\angle EDC = \angle BDA$, $\angle DCE = 2\cdot\angle DEC = 2\alpha$. Since, $\angle BCE = \angle BAE = 2\alpha$, $B,A,C,E$ are concyclic. Let $H$ be foot of altitude from $B$ to $EC$. Let $BH$ and $AE$ meet at $F$. Since $\angle BAC = 120^\circ$, $\angle BEH = 60^\circ$. Let $G$ be on $EC$ such that $\triangle EBG$ is equilateral. Let $BG$ meet $AE$ at $I$. Since $F$ is on the angle bisector of $\triangle EBG$, $EF=FG$ and $\angle IFG = 2\angle FEG = 2\alpha$. So $FG \parallel AB$. By Sin Law, $BI/IG = AB/AC$. If you want synthetic proof: $EI$ cuts $(BEG)$ at $J$. $\triangle JBG \sim \triangle ABC$. $JI$ is angle bisector of $\angle BJG$. So $AB/AC=BJ/JG=BI/IG$ $AB \parallel FG \Rightarrow AB/FG = BI/IG = AB/AC \Rightarrow FG = AC = EF$. We have $\angle EFH = \angle BFA = 90^\circ - \angle AEC = 90^\circ - \alpha$ and also $\angle BAF = 2\alpha$. So $AB=AF$. $AE=AF+FE=AB+AC$. $\blacksquare$