AUT_ABooklet_2020

We shall show that $AB=BX$ holds. Since $AB=AY$ then follows by the same argument, this completes the proof (see Figure 3).

Figure 3: Problem 10

In this solution, we use oriented angles between lines (modulo $180^{\circ }$) with the notation $\angle PQR$. As usual the angles of the triangle $ABC$ are denoted by $\alpha =\angle BAC$, $\beta =\angle CBA$ and $\gamma =\angle ACB$.

The inscribed angle theorem gives $$\angle AXB=\angle AXC=\angle AIC=-\angle CIA=180^{\circ }-\angle CIA=\angle IAC+\angle ACI=\frac{1}{2}(\alpha +\gamma ).$$ This immediately implies $$\angle BAX=-\angle AXB-\angle XBA=-\frac{1}{2}(\alpha +\gamma )-\beta =\frac{1}{2}(\alpha +\gamma ).$$ Therefore, the triangle ABX is indeed isosceles, and we are done.