2024 USAMO 第 5 题

Point $D$ is selected inside acute triangle $ABC$ so that $\angle DAC=\angle ACB$ and $\angle BDC=90^\circ+\angle BAC$. Point $E$ is chosen on ray $BD$ so that $AE=EC$. Let $M$ be the midpoint of $BC$. Show that line $AB$ is tangent to the circumcircle of triangle $BEM$.

点$D$选择在锐角三角形$ABC$内，使得$\angle DAC=\angle ACB$和$\angle BDC=90^\circ+\angle BAC$。在射线$BD$上选择点$E$，使得$AE=EC$。令 $M$ 为 $BC$ 的中点。证明直线 $AB$ 与三角形 $BEM$ 的外接圆相切。