SOLUTION: In triangle ABC, the orthocenter H lies on altitude \overline{AD}. Find \frac{AH}{HD}.

Question 1210491: In triangle ABC, the orthocenter H lies on altitude \overline{AD}. Find \frac{AH}{HD}.
Answer by CPhill(2136) (Show Source): You can put this solution on YOUR website!
The ratio $\frac{AH}{HD}$ is $\mathbf{2}$.
Here is the reasoning based on the properties of an equilateral triangle:
## Orthocenter and Centroid in an Equilateral Triangle
In any triangle, the **orthocenter** ($H$) is the intersection point of the altitudes. The **centroid** ($G$) is the intersection point of the medians.
1. **Coincident Centers:** In an **equilateral triangle** ($\triangle ABC$):
* The median from any vertex is also the altitude from that vertex.
* This means the orthocenter ($H$), the centroid ($G$), the circumcenter, and the incenter all coincide at the same point.
* Therefore, the orthocenter $H$ is the same point as the centroid $G$.
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 $\mathbf{2:1}$.
[Image of a triangle and its medial triangle with area ratio labeled]

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 ($\overline{BC}$).
$$AH = 2 \cdot HD$$
Therefore, the ratio is:
$$\frac{AH}{HD} = \mathbf{2}$$
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