Twice angle conditions with a point in a triangle

Question

Let $X$ be a point lying in the interior of the acute triangle $ABC$ such that $\angle BAX = 2\angle XBA$ and $\angle XAC = 2\angle ACX$.

Denote by $M$ the midpoint of the arc $BC$ of the circumcircle $(ABC)$ containing $A$. Prove that $XM = XA$.

Answer 1

We use symbols on the picture. So we have to show $x=m$

By Ptolomy theorem we have $$ma+cs = bs\implies m = s{b-c\over a}$$

Since $MB = MC =s$ we have $${a\over 2s} = \sin (\alpha+\beta)$$ and if we use Sin theorem for triangles $AXB$ and $AXC$ we get:

$$ {x\over \sin \beta} ={b\over \sin 3\beta}\;\;\;\;\wedge \;\;\;\;{x\over \sin \alpha} ={c\over \sin 3\alpha}$$

so $$b-c = x\Big({\sin 3\beta \over \sin \beta}- {\sin 3\alpha \over \sin \alpha}\Big)$$

So $$ m= x\Big({\sin 3\beta \over \sin \beta}- {\sin 3\alpha \over \sin \alpha}\Big)\cdot {1\over 2\sin (\alpha+\beta)}$$

So you have to check if $$\boxed{2\sin (\alpha+\beta)(\sin 3\beta \sin \alpha - \sin 3\alpha \sin \beta) = \sin\alpha\sin \beta}$$

Can you do it?