ABC is a triangle right angled at B. The hypotenuse (AC) is four times the perpendicular (BD) drawn to it from the opposite vertex and `ADltDC`.
What is AD: DC equal to?

Options

A. A. `(7-2sqrt3):1`
B. B. `(7-4sqrt3):1`
C. C. `1:2`
D. D. none of these

Answer

To solve the problem, we need to find the ratio of segments AD and DC in triangle ABC, which is right-angled at B. We know that the hypotenuse AC is four times the length of the perpendicular BD drawn from vertex B to side AC.

### Step-by-Step Solution:

1. **Define Variables**:
- Let \( BD = X \) (the length of the perpendicular).
- Since \( AC \) (the hypotenuse) is four times \( BD \), we have \( AC = 4X \).

2. **Segment Definitions**:
- Let \( AD = Y \).
- Then, since \( D \) divides \( AC \) into two segments, \( DC = AC - AD = 4X - Y \).

3. **Identify Angles**:
- Let \( \angle A = \theta \).
- Since triangle ABC is right-angled at B, \( \angle B = 90^\circ \).
- Thus, \( \angle C = 90^\circ - \theta \).

4. **Using Trigonometric Ratios**:
- In triangle ABD, we can use the tangent function:
\[ \tan(\theta) = \frac{BD}{AD} = \frac{X}{Y} \quad \text{(1)} \]
- In triangle BDC, we can also use the tangent function:
\[ \tan(\theta) = \frac{BD}{DC} = \frac{X}{4X - Y} \quad \text{(2)} \]

5. **Equating the Two Expressions**:
- Since both expressions equal \( \tan(\theta) \), we can set them equal to each other:
\[ \frac{X}{Y} = \frac{X}{4X - Y} \]

6. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ X(4X - Y) = XY \]
- Expanding this, we have:
\[ 4X^2 - XY = XY \]
- Rearranging gives:
\[ 4X^2 = 2XY \quad \Rightarrow \quad 2X^2 = XY \quad \Rightarrow \quad Y = \frac{2X^2}{X} = 2X \quad \text{(3)} \]

7. **Finding DC**:
- Substitute \( Y \) back into the expression for \( DC \):
\[ DC = 4X - Y = 4X - 2X = 2X \]

8. **Finding the Ratio**:
- Now we have \( AD = Y = 2X \) and \( DC = 2X \).
- Therefore, the ratio \( AD:DC \) is:
\[ AD:DC = 2X:2X = 1:1 \]

### Final Answer:
The ratio \( AD:DC \) is \( 1:1 \).