In a `Delta ABC, angle BAC = 90^@ , AD` is drawn perpendicular from A on BC. Which among the following is the mean proportional between BD and BC?

Options

A. AD
B. AB
C. CD
D. AC

Answer

To find the mean proportional between \( BD \) and \( BC \) in triangle \( ABC \) where \( \angle BAC = 90^\circ \) and \( AD \) is drawn perpendicular from \( A \) to \( BC \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Right Triangle**:
In triangle \( ABC \), since \( \angle BAC = 90^\circ \), we have a right triangle with \( AB \) as one leg, \( AC \) as the other leg, and \( BC \) as the hypotenuse.

**Hint**: Remember that in a right triangle, the sides are related through the Pythagorean theorem.

2. **Draw the Perpendicular**:
Draw the perpendicular \( AD \) from point \( A \) to line \( BC \). This creates two smaller triangles: \( ABD \) and \( ACD \).

**Hint**: Visualizing the triangles helps in understanding the relationships between the sides.

3. **Use Similar Triangles**:
Triangles \( ABD \) and \( ACD \) are similar to triangle \( ABC \) because they all share angle \( A \) and both have a right angle.

**Hint**: Recall that if two triangles have two angles equal, they are similar.

4. **Set Up Proportions**:
From the similarity of triangles, we can set up the following proportion:
\[ \frac{BD}{BC} = \frac{AB}{AC} \]

**Hint**: When dealing with similar triangles, the ratios of corresponding sides are equal.

5. **Cross-Multiply**:
Rearranging the proportion gives us:
\[ BD \cdot AC = AB \cdot BC \]

**Hint**: Cross-multiplying helps isolate the terms we need.

6. **Solve for Mean Proportional**:
To find the mean proportional \( x \) between \( BD \) and \( BC \), we can express it as:
\[ BD^2 = AB \cdot BC \]
Thus, the mean proportional \( BD \) is given by:
\[ BD = \sqrt{AB \cdot BC} \]

**Hint**: The mean proportional is the square root of the product of the two quantities.

### Final Answer:
The mean proportional between \( BD \) and \( BC \) is \( \sqrt{AB \cdot BC} \).