Hellenic Mathematical Olympiad ARCHIMEDES

(α) If $Z$ is the midpoint of $B\Gamma$, then $AZ$ is altitude and median of the isosceles triangle $AB\Gamma$. Then the right angled triangles $A\Gamma Z$ and $A\Gamma \Delta$ are equal, because they have: $A\Gamma$ common hypotenuse and $A\Gamma \Delta =A\Gamma Z$, (the last equality arises from the equalities $A\Gamma \Delta =A\beta \Gamma$ (chord-tangent angle and the corresponding inscribed angle) and $A\beta \Gamma =A\Gamma Z$, (since $AB=A\Gamma$). Hence we get: $$\Gamma \Delta =\Gamma Z=\frac{B\Gamma }{2}.$$

(β) Let $AB<A\Gamma$. We consider the perpendicular bisector of $B\Gamma$ ($M$ is the midpoint) which intersects the extension of $AB$ at $E$. Then $A\beta \Gamma =A\Gamma \Delta$ and from hypothesis we have $2\Gamma \Delta =B\Gamma =2BM$. Hence $\Gamma \Delta =BM$. Therefore the right angled triangles $EBM$ and $A\Gamma \Delta$ are equal, and so $A\Gamma =EB$. Since $EB=E\Gamma$, we conclude $A\Gamma =E\Gamma$. However this is absurd, because $EA\Gamma =180^{\circ }-A$ is obtuse angle and so the triangle $EA\Gamma$ would be having two obtuse angles. Hence $AB=A\Gamma$ and the triangle $AB\Gamma$ is isosceles.