生成解题视频---**Textual Information:** **Question Stem:** 如图, 在 $\triangle ABC$ 中, $AB=AC$, $\angle BAC=2\alpha$ ($0^\circ < \alpha < 90^\circ$), 点 $D$ 为边 $BC$ 上一点 ($BD>CD$), 连接 $AD$, 将线段 $AD$ 绕点 $A$ 逆时针旋转 $2\alpha$ 得到线段 $AE$, 连接 $ED$ 交 $AC$ 于点 $F$, 连接 $CE$. 若点 $M, N, H$ 分别为 $BC, DE, DF$ 的中点, 连接 $MH, NH$, 证明: $MH=NH$. **Translation of Question Stem:** As shown in the figure, in $\triangle ABC$, $AB=AC$, $\angle BAC=2\alpha$ ($0^\circ < \alpha < 90^\circ$), point $D$ is a point on side $BC$ ($BD>CD$). Connect $AD$. Rotate line segment $AD$ counterclockwise around point $A$ by $2\alpha$ to get line segment $AE$. Connect $ED$ which intersects $AC$ at point $F$. Connect $CE$. If points $M, N, H$ are the midpoints of $BC, DE, DF$ respectively, connect $MH, NH$, prove: $MH=NH$. **Diagram Description:** * **Type:** Geometric figure illustrating a triangle and construction lines. * **Elements:** * **Points:** A, B, C, D, E, F, M, N, H. * **Triangles:** $\triangle ABC$, $\triangle ADE$, $\triangle FDC$, $\triangle AFC$, $\triangle AB D$, $\triangle ACD$, $\triangle ACE$, $\triangle ADE$, $\triangle MNH$. * **Lines/Segments:** AB, AC, BC, AD, AE, ED, DC, BD, CE, DE, AC, DF, MH, NH. * **Specific Points:** * D is on BC, with BD > CD. * F is the intersection of ED and AC. * M is the midpoint of BC. * N is the midpoint of DE. * H is the midpoint of DF. * **Relationships/Annotations:** * $\triangle ABC$ is an isosceles triangle with AB = AC. * $\angle BAC = 2\alpha$. * Segment AE is obtained by rotating segment AD counterclockwise around A by angle $2\alpha$. This implies AD = AE and $\angle DAE = 2\alpha$. * Lines MH and NH are drawn, connecting midpoints. * **Labels:** Points are labeled with capital letters A, B, C, D, E, F, M, N, H.