In a triangle ABC, `angleA=30^@` H is the orthocentre and D is the midpoint of `BbarC`. Segment HD is produced to T such that HD=DT. The ratio `(AT)/(DC)` equals to

Options

A. `1:2`
B. `2:1`
C. `3:2`
D. `2:3`

Answer

To solve the problem step by step, we will analyze the triangle ABC with given conditions and find the ratio \( \frac{AT}{DC} \).

### Step 1: Understand the Geometry
In triangle ABC, we know that:
- \( \angle A = 30^\circ \)
- H is the orthocenter of triangle ABC.
- D is the midpoint of segment BC.
- Segment HD is extended to point T such that \( HD = DT \).

### Step 2: Position Vectors
Let:
- \( \vec{A} \) be the position vector of point A.
- \( \vec{B} \) be the position vector of point B.
- \( \vec{C} \) be the position vector of point C.
- \( \vec{D} \) be the position vector of point D, which is the midpoint of BC:
\[ \vec{D} = \frac{\vec{B} + \vec{C}}{2} \]

### Step 3: Position Vector of H
The position vector of the orthocenter H can be expressed as:
\[ \vec{H} = \vec{A} + \vec{B} + \vec{C} - \vec{O} \]
where O is the circumcenter. However, we can also express it in terms of the circumradius \( R \) and the angles of the triangle.

### Step 4: Finding T
Since \( HD = DT \), we can express the position vector of T as:
\[ \vec{T} = \vec{H} + (\vec{D} - \vec{H}) = 2\vec{D} - \vec{H} \]

### Step 5: Calculate Lengths
We need to find the lengths \( AT \) and \( DC \):
1. **Length \( AT \)**:
- Using the properties of the circumradius and the sine rule:
\[ BC = 2R \sin A = 2R \sin 30^\circ = R \]
- Since \( AT \) is twice the length of \( DC \) (because D is the midpoint), we have:
\[ AT = 2 \cdot DC \]

2. **Length \( DC \)**:
- Since D is the midpoint of BC, we can say:
\[ DC = \frac{BC}{2} = \frac{R}{2} \]

### Step 6: Ratio Calculation
Now we can find the ratio:
\[ \frac{AT}{DC} = \frac{2 \cdot DC}{DC} = \frac{2 \cdot \frac{R}{2}}{\frac{R}{2}} = 2 \]

### Final Answer
Thus, the ratio \( \frac{AT}{DC} \) is:
\[ \frac{AT}{DC} = 2 \]