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The ratio $\frac{AH}{HD}$ is $\mathbf{2}$.\r
\n" ); document.write( "\n" ); document.write( "Here is the reasoning based on the properties of an equilateral triangle:\r
\n" ); document.write( "\n" ); document.write( "## Orthocenter and Centroid in an Equilateral Triangle\r
\n" ); document.write( "\n" ); document.write( "In any triangle, the **orthocenter** ($H$) is the intersection point of the altitudes. The **centroid** ($G$) is the intersection point of the medians.\r
\n" ); document.write( "\n" ); document.write( "1. **Coincident Centers:** In an **equilateral triangle** ($\triangle ABC$):
\n" ); document.write( " * The median from any vertex is also the altitude from that vertex.
\n" ); document.write( " * This means the orthocenter ($H$), the centroid ($G$), the circumcenter, and the incenter all coincide at the same point.
\n" ); document.write( " * Therefore, the orthocenter $H$ is the same point as the centroid $G$.\r
\n" ); document.write( "\n" ); document.write( "2. **Median Property:** $\overline{AD}$ is an altitude (given) and, because the triangle is equilateral, it is also a median. The centroid ($G$ or $H$) divides any median into two segments in a ratio of $\mathbf{2:1}$. \r
\n" ); document.write( "\n" ); document.write( "[Image of a triangle and its medial triangle with area ratio labeled]\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "3. **Ratio Calculation:** The segment connecting the vertex ($A$) to the centroid ($H$) is twice as long as the segment connecting the centroid ($H$) to the midpoint ($D$) of the opposite side ($\overline{BC}$).\r
\n" ); document.write( "\n" ); document.write( "$$AH = 2 \cdot HD$$\r
\n" ); document.write( "\n" ); document.write( "Therefore, the ratio is:
\n" ); document.write( "$$\frac{AH}{HD} = \mathbf{2}$$ \n" ); document.write( "